Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Miscellaneous Exercise 7 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7

(I) Select the correct option from the given alternatives.

Question 1.

The line y = mx + 1 is a tangent to the parabola y^{2} = 4x, if m is ________

(A) 1

(B) 2

(C) 3

(D) 4

Answer:

(A) 1

Hint:

y^{2} = 4x

Compare with y^{2} = 4ax

∴ a = 1

Equation of tangent is y = mx + 1

Compare with y = mx + \(\frac{a}{m}\)

\(\frac{a}{m}\) = 1

∴ a = m = 1

Question 2.

The length of latus rectum of the parabola x^{2} – 4x – 8y + 12 = 0 is ________

(A) 4

(B) 6

(C) 8

(D) 10

Answer:

(C) 8

Hint:

Given equation of parabola is

x^{2} – 4x – 8y + 12 = 0

⇒ x^{2} – 4x = 8y – 12

⇒ x^{2} – 4x + 4 = 8y – 12 + 4

⇒ (x – 2)^{2} = 8(y – 1)

Comparing this equation with (x – h)^{2} = 4b(y – k), we get

4b = 8

∴ Length of latus rectum = 4b = 8

Question 3.

If the focus of the parabola is (0, -3), its directrix is y = 3, then its equation is ________

(A) x^{2} = -12y

(B) x^{2} = 12y

(C) y^{2} = 12x

(D) y^{2} = -12x

Answer:

(A) x^{2} = -12y

Hint:

SP^{2} = PM^{2}

⇒ (x – 0)^{2} + (y + 3)^{2} = \(\left|\frac{y-3}{\sqrt{1}}\right|^{2}\)

⇒ x^{2} + y^{2} + 6y + 9 = y^{2} – 6y + 9

⇒ x^{2} = -12y

Question 4.

The co-ordinates of a point on the parabola y^{2} = 8x whose focal distance is 4 are ________

(A) (\(\frac{1}{2}\), ±2)

(B) (1, ±2√2)

(C) (2, ±4)

(D) none of these

Answer:

(C) (2, ±4)

Question 5.

The end points of latus rectum of the parabola y^{2} = 24x are ________

(A) (6, ±12)

(B) (12, ±6)

(C) (6, ±6)

(D) none of these

Answer:

(A) (6, ±12)

Question 6.

Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________

(A) y^{2} = 8x

(B) y^{2} = 32x

(C) y^{2} = 16x

(D) x^{2} = 32y

Answer:

(B) y^{2} = 32x

Hint:

Since directrix is parallel to Y-axis,

The X-axis is the axis of the parabola.

Let the equation of parabola be y^{2} = 4ax.

Equation of directrix is x + 8 = 0

∴ a = 8

∴ required equation of parabola is y^{2} = 32x

Question 7.

The area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the endpoints of its latus rectum is ________

(A) 22 sq. units

(B) 20 sq. units

(C) 18 sq. units

(D) 14 sq. units

Answer:

(C) 18 sq. units

Hint:

x^{2} = 12y

4b = 12

b = 3

Area of triangle = \(\frac{1}{2}\) × AB × OS

= \(\frac{1}{2}\) × 4a × a

= \(\frac{1}{2}\) × 12 × 3

= 18 sq. units

Question 8.

If P(\(\frac{\pi}{4}\)) is any point on the ellipse 9x^{2} + 25y^{2} = 225, S and S’ are its foci, then SP . S’P = ________

(A) 13

(B) 14

(C) 17

(D) 19

Answer:

(C) 17

Hint:

9x^{2} + 25y^{2} = 225

\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

Here, a = 5, b = 3

Eccentricity (e) = \(\frac{4}{5}\)

∴ \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{5}{\left(\frac{4}{5}\right)}=\frac{25}{4}\)

Coordinates of foci are S(4, 0) and S'(-4, 0)

P(θ) = (a cos θ, b sin θ)

Question 9.

The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________

(A) y^{2} = 4x

(B) y^{2} = 8x

(C) y^{2} = -16x

(D) x^{2} = 8y

Answer:

(B) y^{2} = 8x

Hint:

The given points lie in the 1st and 4th quadrants.

∴ Equation of the parabola is y^{2} = 4ax

End points of latus rectum are (a, 2a) and (a, -2a)

∴ a = 2

∴ required equation of parabola is y = 8x

Question 10.

If the parabola y^{2} = 4ax passes through (3, 2), then the length of its latus rectum is ________

(A) \(\frac{2}{3}\)

(B) \(\frac{4}{3}\)

(C) \(\frac{1}{3}\)

(D) 4

Answer:

(B) \(\frac{4}{3}\)

Hint:

Length of latus rectum = 4a

The given parabola passes through (3, 2)

∴ (2)^{2} = 4a(3)

∴ 4a = \(\frac{4}{3}\)

Question 11.

The eccentricity of rectangular hyperbola is

(A) \(\frac{1}{2}\)

(B) \(\frac{1}{2^{\frac{1}{2}}}\)

(C) \(2^{\frac{1}{2}}\)

(D) \(\frac{1}{3^{\frac{1}{2}}}\)

Answer:

(C) \(2^{\frac{1}{2}}\)

Question 12.

The equation of the ellipse having one of the foci at (4, 0) and eccentricity \(\frac{1}{3}\) is

(A) 9x^{2} + 16y^{2} = 144

(B) 144x^{2} + 9y^{2} = 1296

(C) 128x^{2} + 144y^{2} = 18432

(D) 144x^{2} + 128y^{2} = 18432

Answer:

(C) 128x^{2} + 144y^{2} = 18432

Question 13.

The equation of the ellipse having eccentricity \(\frac{\sqrt{3}}{2}\) and passing through (-8, 3) is

(A) 4x^{2} + y^{2} = 4

(B) x^{2} + 4y^{2} = 100

(C) 4x^{2} + y^{2} = 100

(D) x^{2} + 4y^{2} = 4

Answer:

(B) x^{2} + 4y^{2} = 100

Question 14.

If the line 4x – 3y + k = 0 touches the ellipse 5x^{2} + 9y^{2} = 45, then the value of k is

(A) 21

(B) ±3√21

(C) 3

(D) 3(21)

Answer:

(B) ±3√21

Question 15.

The equation of the ellipse is 16x^{2} + 25y^{2} = 400. The equations of the tangents making an angle of 180° with the major axis are

(A) x = 4

(B) y = ±4

(C) x = -4

(D) x = ±5

Answer:

(B) y = ±4

Question 16.

The equation of the tangent to the ellipse 4x^{2} + 9y^{2} = 36 which is perpendicular to 3x + 4y = 17 is

(A) y = 4x + 6

(B) 3y + 4x = 6

(C) 3y = 4x + 6√5

(D) 3y = x + 25

Answer:

(C) 3y = 4x + 6√5

Question 17.

Eccentricity of the hyperbola 16x^{2} – 3y^{2} – 32x – 12y – 44 = 0 is

(A) \(\sqrt{\frac{17}{3}}\)

(B) \(\sqrt{\frac{19}{3}}\)

(C) \(\frac{\sqrt{19}}{3}\)

(D) \(\frac{\sqrt{17}}{3}\)

Answer:

(B) \(\sqrt{\frac{19}{3}}\)

Hint:

16x^{2} – 3y^{2} – 32x – 12y – 44 = 0

⇒ 16(x – 1)^{2} – 3(y + 2)^{2} = 48

⇒ \(\frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1\)

Here, a^{2} = 3 and b^{2} = 16

\(e=\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{3+16}}{\sqrt{3}}=\sqrt{\frac{19}{3}}\)

Question 18.

Centre of the ellipse 9x^{2} + 5y^{2} – 36x – 50y – 164 = 0 is at

(A) (2, 5)

(B) (1, -2)

(C) (-2, 1)

(D) (0, 0)

Answer:

(A) (2, 5)

Hint:

9x^{2} + 5y^{2} – 36x – 50y – 164 = 0

⇒ 9(x – 2)^{2} + 5(y – 5)^{2} = 325

⇒ \(\frac{(x-2)^{2}}{\frac{325}{9}}+\frac{(y-5)^{2}}{65}=1\)

⇒ centre of the ellipse = (2, 5)

Question 19.

If the line 2x – y = 4 touches the hyperbola 4x^{2} – 3y^{2} = 24, the point of contact is

(A) (1, 2)

(B) (2, 3)

(C) (3, 2)

(D) (-2, -3)

Answer:

(C) (3, 2)

Question 20.

The foci of hyperbola 4x^{2} – 9y^{2} – 36 = 0 are

(A) (±√13, 0)

(B) (±√11, 0)

(C) (±√12, 0)

(D) (0, ±√12)

Answer:

(A) (±√13, 0)

II. Answer the following.

Question 1.

For each of the following parabolas, find focus, equation of file directrix, length of the latus rectum and ends of the latus rectum.

(i) If 2y^{2} = 17x

(ii) 5x^{2} = 24y

Solution:

(i) Given equation of the parabola is 2y^{2} = 17x

y^{2} = \(\frac{17}{2}\)x

Comparing this equation with y^{2} = 4ax, we get

4a = \(\frac{17}{2}\)

a = \(\frac{17}{8}\)

Co-ordinates of focus are S(a, 0), i.e., S(\(\frac{17}{8}\), 0)

Equation of the directrix is x + a = 0

x + \(\frac{17}{8}\) = 0

8x + 17 = 0

Length of latus rectum = 4a = 4(\(\frac{17}{8}\)) = \(\frac{17}{2}\)

Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a)

i.e., \(\left(\frac{17}{8}, \frac{17}{4}\right)\) and \(\left(\frac{17}{8},-\frac{17}{4}\right)\)

(ii) Given equation of the parabola is 5x^{2} = 24y

x^{2} = \(\frac{24 y}{5}\)

Comparing this equation with x^{2} = 4by, we get

4b = \(\frac{24}{5}\)

b = \(\frac{6}{5}\)

Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{6}{5}\))

Equation of the directrix is y + b = 0

y + \(\frac{6}{5}\) = 0

5y + 6 = 0

Length of latus rectum = 4b = 4(\(\frac{6}{5}\)) = \(\frac{24}{5}\)

Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b), i.e., \(\left(\frac{12}{5}, \frac{6}{5}\right)\) and \(\left(\frac{-12}{5}, \frac{6}{5}\right)\)

Question 2.

Find the cartesian co-ordinates of the points on the parabola y^{2} = 12x whose parameters are

(i) 2

(ii) -3

Solution:

Given equation of the parabola is y^{2} = 12x

Comparing this equation with y^{2} = 4ax, we get

4a = 12

∴ a = 3

If t is the parameter of the point P on the parabola, then

P(t) = (at^{2}, 2at)

i.e., x = at^{2} and y = 2at …..(i)

(i) Given, t = 2

Substituting a = 3 and t = 2 in (i), we get

x = 3(2)^{2} and y = 2(3)(2)

x = 12 and y = 12

∴ The cartesian co-ordinates of the point on the parabola are (12, 12).

(ii) Given, t = -3

Substitùting a = 3 and t = -3 in (i), we get

x = 3(-3)^{2} and y = 2(3)(-3)

∴ x = 27 and y = -18

∴ The cartesian co-ordinates of the point on the parabola are (27, -18).

Question 3.

Find the co-ordinates of a point of the parabola y^{2} = 8x having focal distance 10.

Solution:

Given equation of the parabola is y^{2} = 8x

Comparing this equation with y^{2} = 4ax, we get

4a = 8

∴ a = 2

Focal distance of a point = x + a

Given, focal distance = 10

x + 2 = 10

∴ x = 8

Substituting x = 8 in y^{2} = 8x, we get

y^{2} = 8(8)

∴ y = ±8

∴ The co-ordinates of the points on the parabola are (8, 8) and (8, -8).

Question 4.

Find the equation of the tangent to the parabola y^{2} = 9x at the point (4, -6) on it.

Solution:

Given equation of the parabola is y^{2} = 9x

Comparing this equation with y^{2} = 4ax, we get

4a = 9

∴ a = \(\frac{9}{4}\)

Equation of the tangent y^{2} = 4ax at (x_{1}, y_{1}) is yy_{1} = 2a(x + x_{1})

The equation of the tangent at (4, -6) is

y(-6) = 2(\(\frac{9}{4}\))(x + 4)

⇒ -6y = \(\frac{9}{2}\) (x + 4)

⇒ -12y = 9x + 36

⇒ 9x + 12y + 36 = 0

⇒ 3x + 4y + 12 = 0

Question 5.

Find the equation of the tangent to the parabola y^{2} = 8x at t = 1 on it.

Solution:

Given equation of the parabola is y^{2} = 8x

Comparing this equation with y^{2} = 4ax, we get

4a = 8

a = 2

t = 1

Equation of tangent with parameter t is yt = x + at^{2}

∴ The equation of tangent with t = 1 is

y(1) = x + 2(1)^{2}

y = x + 2

∴ x – y + 2 = 0

Question 6.

Find the equations of the tangents to the parabola y^{2} = 9x through the point (4, 10).

Solution:

Given equation of the parabola is y^{2} = 9x

Comparing this equation with y^{2} = 4ax, we get

4a = 9

∴ a = \(\frac{9}{4}\)

Equation of tangent to the parabola y^{2} = 4ax having slope m is

y = mx + \(\frac{a}{m}\)

y = mx + \(\frac{9}{4 m}\)

But, (4, 10) lies on the tangent.

10 = 4m + \(\frac{9}{4 m}\)

⇒ 40m = 16m^{2}+ 9

⇒ 16m^{2} – 40m + 9 = 0

⇒ 16m^{2} – 36m – 4m + 9 = 0

⇒ 4m(4m – 9) – 1(4m – 9) = 0

⇒ (4m – 9) (4m – 1) = 0

⇒ 4m – 9 = 0 or 4m – 1 = 0

⇒ m = \(\frac{9}{4}\) or m = \(\frac{1}{4}\)

These are the slopes of the required tangents.

By slope point form, y – y_{1} = m(x – x_{1}),

the equations of the tangents are

y – 10 = \(\frac{9}{4}\)(x – 4) or y – 10 = \(\frac{1}{4}\)(x – 4)

⇒ 4y – 40 = 9x – 36 or 4y – 40 = x – 4

⇒ 9x – 4y + 4 = 0 or x – 4y + 36 = 0

Question 7.

Show that the two tangents drawn to the parabola y^{2} = 24x from the point (-6, 9) are at the right angle.

Solution:

Given the equation of the parabola is y^{2} = 24x.

Comparing this equation with y^{2} = 4ax, we get

4a = 24

⇒ a = 6

Equation of tangent to the parabola y^{2} = 4ax having slope m is

y = mx + \(\frac{a}{m}\)

⇒ y = mx + \(\frac{6}{m}\)

But, (-6, 9) lies on the tangent

9 = -6m + \(\frac{6}{m}\)

⇒ 9m = -6m^{2} + 6

⇒ 6m^{2} + 9m – 6 = 0

The roots m_{1} and m_{2} of this quadratic equation are the slopes of the tangents.

m_{1}m_{2} = -1

Tangents drawn to the parabola y^{2} = 24x from the point (-6, 9) are at a right angle.

Alternate method:

Comparing the given equation with y^{2} = 4ax, we get

4a = 24

⇒ a = 6

Equation of the directrix is x = -6.

The given point lies on the directrix.

Since tangents are drawn from a point on the directrix are perpendicular,

Tangents drawn to the parabola y^{2} = 24x from the point (-6, 9) are at the right angle.

Question 8.

Find the equation of the tangent to the parabola y^{2} = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.

Solution:

Given the equation of the parabola is y^{2} = 8x.

Comparing this equation with y^{2} = 4ax, we get

4a = 8

a = 2

Slope of the line 2x + 2y + 5 = 0 is -1

Since the tangent is parallel to the given line,

slope of the tangent line is m = -1

Equation of tangent to the parabola y^{2} = 4ax having slope m is y = mx + \(\frac{a}{m}\)

Equation of the tangent is

y = -x + \(\frac{2}{-1}\)

x + y + 2 = 0

Point of contact = \(\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)\)

= \(\left(\frac{2}{(-1)^{2}}, \frac{2(2)}{-1}\right)\)

= (2, -4)

Question 9.

A line touches the circle x^{2} + y^{2} = 2 and the parabola y^{2} = 8x. Show that its equation is y = ±(x + 2).

Solution:

Given equation of the parabola is y^{2} = 8x

Comparing this equation with y^{2} = 4ax, we get

4a = 8

a = 2

Equation of tangent to given parabola with slope m is

y = mx + \(\frac{2}{m}\)

m^{2}x – my + 2 = 0 ….(i)

Equation of the circle is x^{2} + y^{2} = 2

Its centre = (0, 0) and Radius = √2

Line (i) touches the circle.

Length of perpendicular from the centre to the line (i) = radius

⇒ \(\left|\frac{m^{2}(0)-m(0)+2}{\sqrt{m^{4}+m^{2}}}\right|\) = √2

⇒ \(\frac{4}{m^{4}+m^{2}}\) = 2

⇒ m^{4} + m^{2} – 2 – 0

⇒ (m^{2} + 2)(m^{2} – 1) = 0

Since m^{2} ≠ -2,

m^{2} – 1 = 0

⇒ m = ±1

When m = 1, equation of the tangent is

y = (1)x + \(\frac{2}{(1)}\)

y = (x + 2) …..(i)

When m = -1, equation of the tangent is

y = (-1)x + \(\frac{2}{(-1)}\)

y = -x – 2

y = -(x + 2) …..(ii)

From (i) and (ii),

equation of the common tangents to the given parabola is y = ±(x + 2)

Question 10.

Two tangents to the parabola y^{2} = 8x meet the tangents at the vertex in P and Q. If PQ = 4, prove that the locus of the point of intersection of the two tangents is y^{2} = 8(x + 2).

Solution:

Given parabola is y^{2} = 8x

Comparing with y^{2} = 4ax, we get,

4a = 8

⇒ a = 2

Let M(t_{1}) and N(t_{2}) be any two points on the parabola.

The equations of tangents at M and N are

yt_{1} = x + \(2 \mathrm{t}_{1}^{2}\) …..(1)

yt_{2} = x + \(2 \mathrm{t}_{2}^{2}\) …(2) ….[∵ a = 2]

Let tangent at M meet the tangent at the vertex in P.

But tangent at the vertex is Y-axis whose equation is x = 0.

⇒ to find P, put x = 0 in (1)

⇒ yt_{1} = \(2 \mathrm{t}_{1}^{2}\)

⇒ y = 2t_{1} …..(t_{1} ≠ 0 otherwise tangent at M will be x = 0)

⇒ P = (0, 2t_{1})

Similarly, Q = (0, 2t_{2})

It is given that PQ = 4

∴ |2t_{1} – 2t_{2}| = 4

∴ |t_{1} – t_{2}| = 2 …..(3)

Let R = (x_{1}, y_{1}) be any point on the required locus.

Then R is the point of intersection of tangents at M and N.

To find R, we solve (1) and (2).

Subtracting (2) from (1), we get

y(t_{1} – t_{2}) = \(2 \mathrm{t}_{1}^{2}-2 \mathrm{t}_{2}^{2}\)

y(t_{1} – t_{2}) = 2(t_{1} – t_{2})(t_{1} + t_{2})

∴ y = 2(t_{1} + t_{2}) …..[∵ M, N are distinct ∴ t_{1} ≠ t_{2}]

i.e., y_{1} = 2(t_{1} + t_{2}) …..(4)

∴ from (1), we get

2t_{1}(t_{1} + t_{2}) = x + \(2 \mathrm{t}_{1}^{2}\)

∴ 2t_{1}t_{2} = x i.e. x_{1} = 2t_{1}t_{2} …..(5)

To find the equation of locus of R(x_{1}, y_{1}),

we eliminate t_{1} and t_{2} from the equations (3), (4) and (5).

We know that,

(t_{1} + t_{2})^{2} = (t_{1} + t_{2})^{2} + 4t_{1}t_{2}

⇒ \(\left(\frac{y_{1}}{2}\right)^{2}=4+4\left(\frac{x_{1}}{2}\right)\) …[By (3), (4) and (5)]

⇒ \(y_{1}^{2}\) = 16 + 8x_{1} = 8(x_{1} + 2)

Replacing x_{1} by x and y_{1} by y,

the equation of required locus is y^{2} = 8(x + 2).

Question 11.

The slopes of the tangents drawn from P to the parabola y^{2} = 4ax are m_{1} and m_{2}, showing that

(i) m_{1} – m_{2} = k

(ii) \(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.

Solution:

Let P(x_{1}, y_{1}) be any point on the parabola y^{2} = 4ax.

Equation of tangent to the parabola y^{2} = 4ax having slope m is y = mx + \(\frac{\mathrm{a}}{\mathrm{m}}\)

This tangent passes through P(x_{1}, y_{1}).

y_{1} = mx_{1} + \(\frac{\mathrm{a}}{\mathrm{m}}\)

my_{1} = m^{2}x_{1} + a

m^{2}x_{1} – my_{1} + a = 0

This is a quadratic equation in ‘m’.

The roots m_{1} and m_{2} of this quadratic equation are the slopes of the tangents drawn from P.

∴ m_{1} + m_{2} = \(\frac{y_{1}}{x_{1}}\), m_{1}m_{2} = \(\frac{a}{x_{1}}\)

Since (x_{1}, y_{1}) and a are constants, m_{1} – m_{2} is a constant.

∴ m_{1} – m_{2} = k, where k is constant.

(ii) Since (x_{1}, y_{1}) and a are constants, m_{1}m_{2} is a constant.

\(\left(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right)\) = k, where k is a constant.

Question 12.

The tangent at point P on the parabola y^{2} = 4ax meets the Y-axis in Q. If S is the focus, show that SP subtends a right angle at Q.

Solution:

Let P(\(a t_{1}^{2}\), 2at_{1}) be a point on the parabola and

S(a, 0) be the focus of parabola y^{2} = 4ax

Since the tangent passing through point P meet Y-axis at point Q,

equation of tangent at P(\(a t_{1}^{2}\), 2at_{1}) is

yt_{1} = x + \(a t_{1}^{2}\) …..(i)

∴ Point Q lie on tangent

∴ put x = 0 in equation (i)

yt_{1} = \(a t_{1}^{2}\)

y = at_{1}

∴ Co-ordinate of point Q(0, at_{1})

S = (a, 0), P(\(a t_{1}^{2}\), 2at_{1}), Q(0, at_{1})

∴ SP subtends a right angle at Q.

Question 13.

Find the

(i) lengths of the principal axes

(ii) co-ordinates of the foci

(iii) equations of directrices

(iv) length of the latus rectum

(v) Distance between foci

(vi) distance between directrices of the curve

(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

(b) 16x^{2} + 25y^{2} = 400

(c) \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)

(d) x^{2} – y^{2} = 16

Solution:

(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 25 and b^{2} = 9

∴ a = 5 and b = 3

Since a > b,

X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10

Length of minor axis = 2b = 2(3) = 6

∴ Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

∴ e = \(\frac{\sqrt{25-9}}{5}\) = \(\frac{4}{5}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0),

i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)

i.e., x = ±\(\frac{5}{\frac{4}{5}}\)

i.e., x = ±\(\frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae = 2 (5) (\(\frac{4}{5}\)) = 8

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(5)}{\frac{4}{5}}\) = \(\frac{25}{2}\)

(b) Given equation of the ellipse is 16x^{2} + 25y^{2} = 400

\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 25 and b^{2} = 16

∴ a = 5 and b = 4

Since a > b,

X-axis is the major axis and Y-axis is the minor axis

(i) Length of major axis = 2a = 2(5) = 10

Length of minor axis = 2b = 2(4) = 8

Lengths of the principal axes are 10 and 8.

(ii) b^{2} = a^{2} (1 – e^{2})

16 = 25(1 – e^{2})

\(\frac{16}{25}\) = 1 – e^{2}

e^{2} = 1 – \(\frac{16}{25}\)

e^{2} = \(\frac{9}{25}\)

e = \(\frac{3}{5}\) ……[∵ 0 < e < 1]

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),

i.e., S(5(\(\frac{3}{5}\)), 0) and S'(-5(\(\frac{3}{5}\)), 0),

i.e., S(3, 0) and S'(-3, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)

i.e., x = ±\(\frac{5}{\left(\frac{3}{5}\right)}\)

i.e., x = ±\(\frac{25}{3}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(16)}{5}=\frac{32}{5}\)

(v) Distance between foci = 2ae = 2(5)(\(\frac{3}{5}\)) = 6

(vi) Distance between directrices = \(\frac{2 a}{e}=\frac{2(5)}{\left(\frac{3}{5}\right)}=\frac{50}{3}\)

(c) Given equation of the hyperbola \(\frac{x^{2}}{144}-\frac{y^{2}}{25}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

a^{2} = 144 and b^{2} = 25

∵ a = 12 and b = 5

(i) Length of transverse axis = 2a = 2(12) = 24

Length of conjugate axis = 2b = 2(5) = 10

lengths of the principal axes are 24 and 10.

(ii) b^{2} = a^{2}(e^{2} – 1)

25 = 144 (e^{2} – 1)

\(\frac{25}{144}\) = e^{2} – 1

e^{2} = 1 + \(\frac{25}{144}\)

e^{2} = \(\frac{169}{144}\)

e = \(\frac{13}{12}\) …….[∵ e > 1]

Co-ordinates of foci are S(ae, 0) and S'(-ae, 0)

i.e., S(12(\(\frac{13}{12}\)), 0) and S'(-12(\(\frac{13}{12}\)), 0)

i.e., S(13, 0) and S'(-13, 0)

(iii) Equations of the directrices are x = \(\pm \frac{a}{e}\)

i.e., x = \(\pm \frac{12}{\left(\frac{13}{12}\right)}\)

i.e., x = \(\pm \frac{144}{13}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(25)}{12}=\frac{25}{6}\)

(v) Distance between foci = 2ae = 2(12)(\(\frac{13}{12}\)) = 26

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{2(12)}{\left(\frac{13}{12}\right)}\) = \(\frac{288}{13}\)

(d) Given equation of the hyperbola is x^{2} – y^{2} = 16

∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{16}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 16 and b^{2} = 16

∴ a = 4 and b = 4

(i) Length of transverse axis = 2a = 2(4) = 8

Length of conjugate axis = 2b = 2(4) = 8

(ii) We know that

Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),

i.e., S(4√2, 0) and S'(-4√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\)

∴x = ± \(\frac{4}{\sqrt{2}}\)

∴ x = ±2√2

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac{2(16)}{4}\) = 8

(v) Distance between foci = 2ae = 2(4)(√2) = 8√2

(vi) Distance between directrices = \(\frac{2 a}{e}\) = \(\frac{2(4)}{\sqrt{2}}\) = 4√2.

Question 14.

Find the equation of the ellipse in standard form if

(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.

(ii) the length of the major axis is 10 and the distance between foci is 8.

(iii) passing through the points (-3, 1) and (2, -2).

Solution:

(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

Given, eccentricity (e) = \(\frac{3}{8}\)

Distance between foci = 2ae

Given, distance between foci = 6

∴ 2ae = 6

∴ 2a(\(\frac{3}{8}\)) = 6

∴ \(\frac{3a}{4}\) = 6

∴ a = 8

∴ a^{2} = 64

Now, b^{2} = a^{2} (1 – e^{2})

= \(64\left[1-\left(\frac{3}{8}\right)^{2}\right]\)

= \(4\left(1-\frac{9}{64}\right)\)

= 64(\(\frac{55}{64}\))

= 55

∴ The required equation of the ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\)

(ii) Let the equation of the ellipse be

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) ……(1)

Then length of major axis = 2a = 10

∴ a = 5

Also, distance between foci= 2ae = 8

∴ 2 × 5 × e = 8

∴ e = \(\frac{4}{5}\)

∴ b^{2} = a^{2}(1 – e^{2})

= 25(1 – \(\frac{6}{25}\))

= 9

∴ from (1), the equation of the required ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.

The ellipse passes through the points (-3, 1) and (2, -2).

∴ Substituting x = -3 and y = 1 in equation of ellipse, we get

\(\frac{(-3)^{2}}{a^{2}}+\frac{1^{2}}{b^{2}}=1\)

∴ \(\frac{9}{a^{2}}+\frac{1}{b^{2}}=1\) …..(i)

Substituting x = 2 and y = -2 in equation of ellipse, we get

\(\frac{2^{2}}{a^{2}}+\frac{(-2)^{2}}{b^{2}}=1\)

∴ \(\frac{4}{a^{2}}+\frac{4}{b^{2}}=1\) ……(ii)

Let \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B

∴ Equations (i) and (ii) become

9A + B = 1 ..…(iii)

4A + 4B = 1 …..(iv)

Multiplying (iii) by 4, we get

36A + 4B = 4 …..(v)

Subtracting (iv) from (v), we get

32A = 3

∴ A = \(\frac{3}{32}\)

Substituting A = \(\frac{3}{32}\) in (iv), we get

4(\(\frac{3}{32}\)) + 4B = 1

∴ \(\frac{3}{8}\) + 4B = 1

∴ 4B = 1 – \(\frac{3}{8}\)

∴ 4B = \(\frac{5}{8}\)

∴ B = \(\frac{5}{32}\)

Since \(\frac{1}{a^{2}}\) = A and \(\frac{1}{b^{2}}\) = B

\(\frac{1}{a^{2}}=\frac{3}{32}\) and \(\frac{1}{b^{2}}=\frac{5}{32}\)

∴ a^{2} = \(\frac{32}{3}\) and b^{2} = \(\frac{32}{5}\)

∴ The required equation of ellipse is

\(\frac{x^{2}}{\left(\frac{32}{3}\right)}+\frac{y^{2}}{\left(\frac{32}{5}\right)}\)

i.e., 3x^{2} + 5y^{2} = 32.

Question 15.

Find the eccentricity of an ellipse if the distance between its directrices is three times the distance between its foci.

Solution:

Let the equation of the ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

It is given that,

distance between directrices is three times the distance between the foci.

∴ \(\frac{2a}{e}\) = 3(2ae)

∴ 1 = 3e^{2}

∴ e^{2} = \(\frac{1}{3}\)

∴ e = \(\frac{1}{\sqrt{3}}\) …..[∵ 0 < e < 1]

Question 16.

For the hyperbola \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\), prove that SA . S’A = 25, where S and S’ are the foci and A is the vertex.

Solution:

Given equation of the hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{25}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 100 and b^{2} = 25

∴ a = 10 and b = 5

∴ Co-ordinates of vertex is A(a, 0), i.e., A(10, 0)

Eccentricity, e = \(\frac{\sqrt{a^{2}+b^{2}}}{a}\)

= \(\frac{\sqrt{100+25}}{10}\)

= \(\frac{\sqrt{125}}{10}\)

= \(\frac{5 \sqrt{5}}{10}\)

= \(\frac{\sqrt{5}}{2}\)

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(10(\(\frac{\sqrt{5}}{2}\)), 0) and S'(-10(\(\frac{\sqrt{5}}{2}\)), 0)

i.e., S(5√5, 0) and S'(-5√5, 0)

Since S, A and S’ lie on the X-axis,

SA = |5√5 – 10| and S’A = |-5√5 – 10|

= |-(5√5 + 10)|

= |5√5 + 10|

∴ SA . S’A = |5√5 – 10| |5√5 + 10|

= |(5√5)^{2} – (10)^{2}|

= |125 – 100|

= |25|

SA . S’A = 25

Question 17.

Find the equation of the tangent to the ellipse \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 5 and b^{2} = 4

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)

Since (2, -2) lies on both the tangents,

-2 = 2m ± \(\sqrt{5 m^{2}+4}\)

∴ -2 – 2m = ±\(\sqrt{5 m^{2}+4}\)

Squaring both the sides, we get

4m^{2} + 8m + 4 = 5m^{2} + 4

∴ m^{2} – 8m = 0

∴ m(m – 8) = 0

∴ m = 0 or m = 8

These are the slopes of the required tangents.

∴ By slope point form y – y_{1} = m(x – x_{1}),

the equations of the tangents are

y + 2 = 0(x – 2) and y + 2 = 8(x – 2)

∴ y + 2 = 0 and y + 2 = 8x – 16

∴ y + 2 = 0 and 8x – y – 18 = 0.

Question 18.

Find the equation of the tangent to the ellipse x^{2} + 4y^{2} = 100 at (8, 3).

Solution:

Given equation of ellipse is x^{2} + 4y^{2} = 100

∴ \(\frac{x^{2}}{100}+\frac{y^{2}}{25}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 100 and b^{2} = 25

Equation of tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at (x_{1}, y_{1}) is \(\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1\)

Equation of tangent at (8, 3) is

\(\frac{8 x}{100}+\frac{3 y}{25}=1\)

\(\frac{2 x}{25}+\frac{3 y}{25}=1\)

2x + 3y = 25

Question 19.

Show that the line 8y + x = 17 touches the ellipse x^{2} + 4y^{2} = 17. Find the point of contact.

Solution:

Question 20.

Tangents are drawn through a point P to the ellipse 4x^{2} + 5y^{2} = 20 having inclinations θ_{1} and θ_{2} such that tan θ_{1} + tan θ_{2} = 2. Find the equation of the locus of P.

Solution:

Given equation of the ellipse is 4x^{2} + 5y^{2} = 20.

∴ \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 5 and b^{2} = 4

Since inclinations of tangents are θ_{1} and θ_{2},

m_{1} = tan θ_{1} and m_{2} = tan θ_{2}

Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)

∴ y = mx ± \(\sqrt{5 m^{2}+4}\)

∴ y – mx = ±\(\sqrt{5 m^{2}+4}\)

Squaring both the sides, we get

y^{2} – 2mxy + m^{2}x^{2} = 5m^{2} + 4

∴ (x^{2} – 5)m^{2} – 2xym + (y^{2} – 4) = 0

The roots m_{1} and m_{2} of this quadratic equation are the slopes of the tangents.

∴ m_{1} + m_{2} = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)

Given, tan θ_{1} + tan θ_{2} = 2

∴ m_{1} + m_{2} = 2

∴ \(\frac{2 x y}{x^{2}-5}\)

∴ xy = x^{2} – 5

∴ x^{2} – xy – 5 = 0, which is the required equation of the locus of P.

Question 21.

Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.

Solution:

Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get

∴ a^{2} = 25, b^{2} = 16

∴ a = 5, b = 4

We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)

∴ e = \(\frac{\sqrt{25-16}}{5}\) = \(\frac{3}{5}\)

ae = 5(\(\frac{3}{5}\)) = 3

Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),

i.e., S(3, 0) and S'(-3, 0)

Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

y = mx ± \(\sqrt{\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}}\)

Equation of one of the tangents to the ellipse is

y = mx + \(\sqrt{25 \mathrm{~m}^{2}+16}\)

∴ mx – y + \(\sqrt{25 \mathrm{~m}^{2}+16}\) = 0 …..(i)

p_{1} = length of perpendicular segment from S(3, 0) to the tangent (i)

p_{2} = length of perpendicular segment from S'(-3, 0) to the tangent (i)

Question 22.

Find the equation of the hyperbola in the standard form if

(i) Length of conjugate axis is 5 and distance between foci is 13.

(ii) eccentricity is \(\frac{3}{2}\) and distance between foci is 12.

(iii) length of the conjugate axis is 3 and the distance between the foci is 5.

Solution:

(i) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

Length of conjugate axis = 2b

Given, length of conjugate axis = 5

2b = 5

b = \(\frac{5}{2}\)

b^{2} = \(\frac{25}{4}\)

Distance between foci = 2ae

Given, distance between foci = 13

2ae = 13

ae = \(\frac{13}{2}\)

a^{2}e^{2} = \(\frac{169}{4}\)

Now, b^{2} = a^{2}(e^{2} – 1)

b^{2} = a^{2}e^{2} – a^{2}

\(\frac{25}{4}\) = \(\frac{169}{4}\) – a^{2}

a^{2} = \(\frac{169}{4}-\frac{25}{4}\) = 36

∴ The required equation of hyperbola is \(\frac{x^{2}}{36}-\frac{y^{2}}{\frac{25}{4}}=1\)

i.e., \(\frac{x^{2}}{36}-\frac{4 y^{2}}{25}=1\)

(ii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

Given, eccentricity (e) = \(\frac{3}{2}\)

Distance between foci = 2ae

Given, distance between foci = 12

∴ 2ae = 12

∴ 2a(\(\frac{3}{2}\)) = 12

∴ 3a = 12

∴ a = 4

∴ a^{2} = 16

Now, b^{2} = a^{2}(e^{2} – 1)

∴ b^{2} = \(\left[\left(\frac{3}{2}\right)^{2}-1\right]\)

∴ b^{2} = 16(\(\frac{9}{4}\) – 1)

∴ b^{2} = 16(\(\frac{5}{4}\))

∴ b^{2} = 20

∴ The required equation of hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{20}=1\)

(iii) Let the required equation of hyperbola be \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

Length of conjugate axis = 2b

Given, length of conjugate axis = 3

∴ 2b = 3

∴ b = \(\frac{3}{2}\)

∴ b^{2} = \(\frac{9}{4}\)

Distance between foci = 2ae

Given, distance between foci = 5

∴ 2ae = 5

∴ ae = \(\frac{5}{2}\)

∴ a^{2}e^{2} = \(\frac{25}{4}\)

Now, b^{2} = a^{2}(e^{2} – 1)

∴ b^{2} = a^{2}e^{2} – a^{2}

∴ \(\frac{9}{4}\) = \(\frac{25}{4}\) – a^{2}

∴ a^{2} = \(\frac{25}{4}-\frac{9}{4}\)

∴ a^{2} = 4

∴ The required equation of hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{\left(\frac{9}{4}\right)}=1\)

i.e., \(\frac{x^{2}}{4}-\frac{4 y^{2}}{9}=1\)

Question 23.

Find the equation of the tangent to the hyperbola,

(i) 7x^{2} – 3y^{2} = 51 at (-3, -2)

(ii) x = 3 sec θ, y = 5 tan θ at θ = π/3

(iii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\) at P(30°).

Solution:

(i) Given equation of the hyperbola is 7x^{2} – 3y^{2} = 51

(ii) Given, equation of the hyperbola is

x = 3 sec θ, y = 5 tan θ

Since sec^{2} θ – tan^{2} θ = 1,

\(\frac{x^{2}}{9}-\frac{y^{2}}{25}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 9 and b^{2} = 25

a = 3 and b = 5

Equation of tangent at P(θ) is

\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)

∴ Equation of tangent at P(π/3) is

\(\frac{x \sec \left(\frac{\pi}{3}\right)}{3}-\frac{y \tan \left(\frac{\pi}{3}\right)}{5}=1\)

\(\frac{2 x}{3}-\frac{\sqrt{3} y}{5}=1\)

10x – 3√3 y = 15

(iii) Given equation of hyperbola is \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 25 and b^{2} = 16

a = 5 and b = 4

Equation of tangent at P(θ) is

\(\frac{x \sec \theta}{\mathrm{a}}-\frac{y \tan \theta}{\mathrm{b}}=1\)

The equation of tangent at P(30°) is

\(\frac{x \sec 30^{\circ}}{5}-\frac{y \tan 30^{\circ}}{4}=1\)

\(\frac{2 x}{5 \sqrt{3}}-\frac{y}{4 \sqrt{3}}=1\)

8x – 5y = 20√3

Question 24.

Show that the line 2x – y = 4 touches the hyperbola 4x^{2} – 3y^{2} = 24. Find the point of contact.

Solution:

Given equation of die hyperbola is 4x^{2} – 3y^{2} = 24.

∴ \(\frac{x^{2}}{6}-\frac{y^{2}}{8}=1\)

Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get

a^{2} = 6 and b^{2} = 8

Given equation of line is 2x – y = 4

∴ y = 2x – 4

Comparing this equation with y = mx + c, we get

m = 2 and c = -4

For the line y = mx + c to be a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we must have

c^{2} = a^{2}m^{2} – b^{2}

c^{2} = (-4)^{2} = 16

a^{2}m^{2} – b^{2} = 6(2)^{2} – 8 = 24 – 8 = 16

∴ The given line is a tangent to the given hyperbola and point of contact

= \(\left(-\frac{\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}},-\frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)

= \(\left(\frac{-6(2)}{-4}, \frac{-8}{-4}\right)\)

= (3, 2)

Question 25.

Find the equations of the tangents to the hyperbola 3x^{2} – y^{2} = 48 which are perpendicular to the line x + 2y – 7 = 0.

Solution:

Given the equation of the hyperbola is 3x^{2} – y^{2} = 48.

∴ \(\frac{x^{2}}{16}-\frac{y^{2}}{48}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 16 and b^{2} = 48

Slope of the line x + 2y – 7 = 0 is \(-\frac{1}{2}\)

Since the given line is perpendicular to the tangents,

slope of the required tangent (m) = 2

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Equations of tangents to the ellipse having slope m are

y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

y = 2x ± \(\sqrt{16(2)^{2}-48}\)

y = 2x ± √16

∴ y = 2x ± 4

Question 26.

Two tangents to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) make angles θ_{1}, θ_{2}, with the transverse axis. Find the locus of their point of intersection if tan θ_{1} + tan θ_{2} = k.

Solution:

Given equation of the hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)

Let θ_{1} and θ_{2} be the inclinations.

m_{1} = tan θ_{1}, m_{2} = tan θ_{2}

Let P(x_{1}, y_{1}) be a point on the hyperbola

Equation of a tangent with slope ‘m’ to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is

y = mx ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

This tangent passes through P(x_{1}, y_{1}).

y_{1} = mx_{1} ± \(\sqrt{a^{2} m^{2}-b^{2}}\)

(y_{1} – mx_{1})^{2} = a^{2}m^{2} – b^{2}

\(\left(x_{1}{ }^{2}-\mathrm{a}^{2}\right) \mathrm{m}^{2}-2 x_{1} y_{1} \mathrm{~m}+\left(y_{1}{ }^{2}+\mathrm{b}^{2}\right)=0\) ……(i)

This is a quadratic equation in ‘m’.

It has two roots say m_{1} and m_{2}, which are the slopes of two tangents drawn from P.

∴ m_{1} + m_{2} = \(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}\)

Since tan θ_{1} + tan θ_{2} = k,

\(\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=k\)

∴ P(x_{1}, y_{1}) moves on the curve whose equation is k(x^{2} – a^{2}) = 2xy.