Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1

Question 1.

Find co-ordinates of focus, equation of directrix, length of latus rectum and the co-ordinates of end points of latus rectum of the parabola:

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

(ii) y^{2} = -20x

(iii) 3x^{2} = 8y

(iv) x^{2} = -8y

(v) 3y^{2} = -16x

Solution:

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

⇒ y^{2} = \(\frac{24}{5}\)x

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

⇒ 4a = \(\frac{24}{5}\)

⇒ a = \(\frac{6}{5}\)

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

Equation of the directrix is x + a = 0.

⇒ x + \(\frac{6}{5}\) = 0

⇒ 5x + 6 = 0

Length of latus rectum = 4a

= 4(\(\frac{6}{5}\))

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

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

⇒ \(\left(\frac{6}{5}, \frac{12}{5}\right)\) and \(\left(\frac{6}{5}, \frac{-12}{5}\right)\)

(ii) Given equation of the parabola is y^{2} = -20x.

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

⇒ 4a = 20

⇒ a = 5

Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)

Equation of the directrix is x – a = 0

⇒ x – 5 = 0

Length of latus rectum = 4a = 4(5) = 20

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

⇒ (-5, 10) and (-5, -10).

(iii) Given equation of the parabola is 3x^{2} = 8y

⇒ x^{2} = \(\frac{8}{3}\) y

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

⇒ 4b = \(\frac{8}{3}\)

⇒ b = \(\frac{2}{3}\)

Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{2}{3}\))

Equation of the directrix is y + b = 0,

⇒ y + \(\frac{2}{3}\) = 0

⇒ 3y + 2 = 0

Length of latus rectum = 4b = 4(\(\frac{2}{3}\)) = \(\frac{8}{3}\)

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

⇒ \(\left(\frac{4}{3}, \frac{2}{3}\right)\) and \(\left(-\frac{4}{3}, \frac{2}{3}\right)\).

(iv) Given equation of the parabola is x^{2} = -8y.

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

⇒ 4b = 8

⇒ b = 2

Co-ordinates of focus are S(0, -b), i.e., S(0, – 2)

Equation of the directrix is y – b = 0, i.e., y – 2 = 0

Length of latus rectum = 4b = 4(2) = 8

∴ Co-ordinates of end points of latus rectum are (2b, -b) and (-2b, -b), i.e., (4, -2) and (-4, -2).

(v) Given equation of the parabola is 3y^{2} = -16x.

⇒ y^{2} = \(-\frac{16}{3}\)x

Comparing this equation withy = -4ax, we get

⇒ 4a = \(\frac{16}{3}\)

⇒ a = \(\frac{4}{3}\)

Co-ordinates of focus are S(-a, 0), i.e., (\(-\frac{4}{3}\), 0)

Equation of the directrix is x – a = 0,

⇒ x – \(-\frac{4}{3}\) = 0

⇒ 3x – 4 = 0

Length of latus rectum = 4a = 4(\(\frac{4}{3}\)) = \(\frac{16}{3}\)

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

i.e., \(\left(-\frac{4}{3}, \frac{8}{3}\right)\) and \(\left(-\frac{4}{3},-\frac{8}{3}\right)\)

Question 2.

Find the equation of the parabola with vertex at the origin, the axis along the Y-axis, and passing through the point (-10, -5).

Solution:

Vertex of the parabola is at origin (0, 0) and its axis is along Y-axis.

Equation of the parabola can be either x^{2} = 4by or x^{2} = -4by

Since the parabola passes through (-10, -5), it lies in 3rd quadrant.

Required parabola is x^{2} = -4by.

Substituting x = -10 and y = -5 in x^{2} = -4by, we get

⇒ (-10)^{2} = -4b(-5)

⇒ b = \(\frac{100}{20}\) = 5

∴ The required equation of the parabola is x^{2} = -4(5)y, i.e., x^{2} = -20y.

Question 3.

Find the equation of the parabola with vertex at the origin, the axis along the X-axis, and passing through the point (3, 4).

Solution:

Vertex of the parabola is at the origin (0, 0) and its axis is along X-axis.

Equation of the parabola can be either y^{2} = 4ax or y^{2} = -4ax.

Since the parabola passes through (3, 4), it lies in the 1st quadrant.

Required parabola is y^{2} = 4ax.

Substituting x = 3 and y = 4 in y^{2} = 4ax, we get

⇒ (4)^{2} = 4a(3)

⇒ a = \(\frac{16}{12}=\frac{4}{3}\)

The required equation of the parabola is

y^{2} = 4(\(\frac{4}{3}\))x

⇒ 3y^{2} = 16x

Question 4.

Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).

Solution:

Focus of the parabola is S(-7, 0) and vertex is O(0, 0).

Since focus lies on X-axis, it is the axis of the parabola.

Focus S(-7, 0) lies on the left-hand side of the origin.

It is a left-handed parabola.

Required parabola is y = -4ax.

Focus is S(-a, 0).

a = 7

∴ The required equation of the parabola is y^{2} =-4(7)x, i.e., y^{2} = -28x.

Question 5.

Find the equation of the parabola with vertex at the origin, the axis along X-axis, and passing through the point

(i) (1, -6)

(ii) (2, 3)

Solution:

(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.

Equation of the parabola can be either y^{2} = 4ax or y^{2} = -4ax.

Since the parabola passes through (1, -6), it lies in the 4th quadrant.

Required parabola is y^{2} = 4ax.

Substituting x = 1 and y = -6 in y^{2} = 4ax, we get

⇒ (-6)^{2} = 4a(1)

⇒ 36 = 4a

⇒ a = 9

∴ The required equation of the parabola is y^{2} = 4(9)x, i.e., y^{2} = 36x.

(ii) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.

Equation of the parabola can be either y^{2} = 4ax or y^{2} = -4ax.

Since the parabola passes through (2, 3), it lies in 1st quadrant.

∴ Required parabola is y^{2} = 4ax.

Substituting x = 2 and y = 3 in y^{2} = 4ax, we get

⇒ (3)^{2} = 4a(2)

⇒ 9 = 8a

⇒ a = \(\frac{9}{8}\)

The required equation of the parabola is

y^{2} = 4(\(\frac{9}{8}\))x

⇒ y^{2} = \(\frac{9}{2}\) x

⇒ 2y^{2} = 9x.

Question 6.

For the parabola 3y^{2} = 16x, find the parameter of the point:

(i) (3, -4)

(ii) (27, -12)

Solution:

Given the equation of the parabola is 3y^{2} = 16x.

⇒ y^{2} = \(\frac{16}{3}\)x

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

⇒ 4a = \(\frac{16}{3}\)

⇒ a = \(\frac{4}{3}\)

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 point is (3, -4)

Substituting x = 3, y = -4 and a = \(\frac{4}{3}\) in (i), we get

3 = \(\frac{4}{3}\) t^{2} and -4 = 2(\(\frac{4}{3}\)) t

∴ The parameter of the given point is \(\frac{-3}{2}\)

(ii) Given point is (27, -12)

Substituting x = 27, y = -12 and a = \(\frac{4}{3}\) in (i), we get

∴ The parameter of the given point is \(\frac{-9}{2}\)

Question 7.

Find the focal distance of a point on the parabola y^{2} = 16x whose ordinate is 2 times the abscissa.

Solution:

Given the equation of the parabola is y^{2} = 16x.

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

⇒ 4a = 16

⇒ a = 4

Since ordinate is 2 times the abscissa,

y = 2x

Substituting y = 2x in y^{2} = 16x, we get

⇒ (2x)^{2} = 16x

⇒ 4x^{2} = 16x

⇒ 4x^{2} – 16x = 0

⇒ 4x(x – 4) = 0

⇒ x = 0 or x = 4

When x = 4,

focal distance = x + a = 4 + 4 = 8

When x = 0,

focal distance = a = 4

∴ Focal distance is 4 or 8.

Question 8.

Find coordinates of the point on the parabola. Also, find focal distance.

(i) y^{2} = 12x whose parameter is \(\frac{1}{3}\)

(ii) 2y^{2} = 7x whose parameter is -2

Solution:

(i) 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)

Given, t = \(\frac{1}{3}\)

Substituting a = 3 and t = \(\frac{1}{3}\) in (i), we get

x = 3(\(\frac{1}{3}\))^{2} and y = 2(3)(\(\frac{1}{3}\))

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

The co-ordinates of the point on the parabola are (\(\frac{1}{3}\), 2)

∴ Focal distance = x + a

= \(\frac{1}{3}\) + 3

= \(\frac{10}{3}\)

(ii) Given equation of the parabola is 2y^{2} = 7x.

⇒ y^{2} = \(\frac{7}{2}\)x

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

⇒ 4a = \(\frac{7}{2}\)

⇒ a = \(\frac{7}{8}\)

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)

Given, t = -2

Substituting a = \(\frac{7}{8}\) and t = -2 in (i), we get

x = \(\frac{7}{8}\)(-2)2 and y = 2(\(\frac{7}{8}\))(-2)

x = \(\frac{7}{2}\) and y = \(\frac{-7}{2}\)

The co-ordinates of the point on the parabola are (\(\frac{7}{2}\), \(\frac{-7}{2}\))

∴ Focal distance = x + a

= \(\frac{7}{2}\) + \(\frac{7}{8}\)

= \(\frac{35}{8}\)

Question 9.

For the parabola y^{2} = 4x, find the coordinates of the point whose focal distance is 17.

Solution:

Given the equation of the parabola is y^{2} = 4x.

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

⇒ 4a = 4

⇒ a = 1

Focal distance of a point = x + a

Given, focal distance = 17

⇒ x + 1 = 17

⇒ x = 16

Substituting x = 16 in y^{2} = 4x, we get

⇒ y^{2} = 4(16)

⇒ y^{2} = 64

⇒ y = ±8

∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).

Question 10.

Find the length of the latus rectum of the parabola y^{2} = 4ax passing through the point (2, -6).

Solution:

Given equation of the parabola is y^{2} = 4ax and it passes through point (2, -6).

Substituting x = 2 and y = -6 in y^{2} = 4ax, we get

⇒ (-6)^{2} = 4a(2)

⇒ 4a = 18

∴ Length of latus rectum = 4a = 18 units

Question 11.

Find the area of the triangle formed by the line joining the vertex of the parabola x^{2} = 12y to the endpoints of the latus rectum.

Solution:

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

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

⇒ 4b = 12

⇒ b = 3

The co-ordinates of focus are S(0, b), i.e., S(0, 3)

End points of the latus-rectum are L(2b, b) and L'(-2b, b),

i.e., L(6, 3) and L'(-6, 3)

Also l(LL’) = length of latus-rectum = 4b = 12

l(OS) = b = 3

Area of ∆OLL’ = \(\frac{1}{2}\) × l(LL’) × l(OS)

= \(\frac{1}{2}\) × 12 × 3

Area of ∆OLL’ = 18 sq. units

Question 12.

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.

Solution:

Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.

It is given that ON = 5 cm and LM = 20 cm.

LN = 10 cm

Taking O as the origin, ON along X-axis and a line through O ⊥ ON as Y-axis.

Let the equation of the reflector be y^{2} = 4ax ……(i)

The point L has the co-ordinates (5, 10) and lies on parabola given by (i).

Substituting x = 5 and y = 10 in (i), we get

⇒ 10^{2} = 4a(5)

⇒ 100 = 20a

⇒ a = 5

Focus is at (a, 0), i.e., (5, 0)

Question 13.

Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x^{2} – 2x + 3.

Solution:

Given equation of the parabola is y = x^{2} – 2x + 3

⇒ y = x^{2} – 2x + 1 + 2

⇒ y – 2 = (x – 1)^{2}

⇒ (x – 1)^{2} = y – 2

Comparing this equation with X^{2} = 4bY, we get

X = x – 1, Y = y – 2

⇒ 4b = 1

⇒ b = \(\frac{1}{4}\)

The co-ordinates of vertex are (X = 0, Y = 0)

⇒ x – 1 = 0 and y – 2 = 0

⇒ x = 1 and y = 2

The co-ordinates of vertex are (1, 2).

The co-ordinates of focus are S(X = 0, Y = b)

⇒ x – 1 = 0 and y – 2 = \(\frac{1}{4}\)

⇒ x = 1 and y = \(\frac{9}{4}\)

The co-ordinates of focus are (1, \(\frac{9}{4}\))

Equation of the axis is X = 0

x – 1 = 0, i.e., x = 1

Equation of directrix is Y + b = 0

⇒ y – 2 + \(\frac{1}{4}\) = 0

⇒ y – \(\frac{7}{4}\) = 0

⇒ 4y – 7 = 0

Question 14.

Find the equation of tangent to the parabola

(i) y^{2} = 12x from the point (2, 5)

(ii) y^{2} = 36x from the point (2, 9)

Solution:

(i) Given equation of the parabola is y^{2} = 12x.

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

⇒ 4a = 12

⇒ a = 3

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

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

Since the tangent passes through the point (2, 5)

⇒ 5 = 2m + \(\frac{3}{m}\)

⇒ 5m = 2m^{2} + 3

⇒ 2m^{2} – 5m + 3 = 0

⇒ 2m^{2} – 2m – 3m + 3 = 0

⇒ 2m(m – 1) – 3(m – 1) = 0

⇒ (m- 1)(2m – 3) = 0

⇒ m = 1 or m = \(\frac{3}{2}\)

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 – 5 = 1(x – 2) and y – 5 = \(\frac{3}{2}\) (x – 2)

⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6

⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0

(ii) Given equation of the parabola is y^{2} = 36x.

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

⇒ 4a = 36

⇒ a = 9

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

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

Since the tangent passes through the point (2, 9),

⇒ 9 = 2m + \(\frac{9}{m}\)

⇒ 9m = 2m^{2} + 9

⇒ 2m^{2} – 9m + 9 = 0

⇒ 2m^{2} – 6m – 3m + 9 = 0

⇒ 2m(m – 3) – 3(m – 3) = 0

⇒ (m – 3)(2m – 3) = 0

⇒ m = 3 or m = \(\frac{3}{2}\)

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 – 9 = 3(x – 2) and y – 9 = \(\frac{3}{2}\) (x – 2)

⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6

⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0

Question 15.

If the tangents drawn from the point (-6, 9) to the parabola y^{2} = kx are perpendicular to each other, find k.

Solution:

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

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

⇒ 4a = k

⇒ a = \(\frac{\mathrm{k}}{4}\)

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

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

Since the tangent passes through the point (-6, 9),

⇒ 9 = -6m + \(\frac{k}{4m}\)

⇒ 36m = -24m2 + k

⇒ 24m^{2} + 36m – k = 0

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

m_{1}m_{2} = \(\frac{-\mathrm{k}}{24}\)

Since the tangents are perpendicular to each other,

m_{1}m_{2} = -1

⇒ \(\frac{-\mathrm{k}}{24}\) = -1

⇒ k = 24

Alternate method:

We know that, tangents drawn from a point on directrix are perpendicular.

(-6, 9) lies on the directrix x = -a.

⇒ -6 = -a

⇒ a = 6

Since 4a = k

⇒ k = 4(6) = 24

Question 16.

Two tangents to the parabola y^{2} = 8x meet the tangents at the vertex in the points P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangents is y^{2} = 8(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 at A(t_{1}) is y

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

Equation of tangent to given parabola at B(t_{2}) is y

t_{2} = x + 2\(\mathrm{t}_{2}^{2}\) …..(ii)

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

x-coordinate of points P and Q is 0.

Let P be(0, k_{1}) and Q be (0, k_{2}).

Then, from (i) and (ii), we get

∴ Equation of locus of R is y^{2} = 8(x + 2).

Question 17.

Find the equation of common tangent to the parabolas y^{2} = 4x and x^{2} = 32y.

Solution:

Given equation of the parabola is y^{2} = 4x

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

⇒ 4a = 4

⇒ a = 1

Let the equation of common tangent be

y = mx + \(\frac{1}{m}\) …..(i)

Substituting y = mx + \(\frac{1}{m}\) in x^{2} = 32y, we get

⇒ x^{2} = 32(mx + \(\frac{1}{m}\)) = 32 mx + \(\frac{32}{m}\)

⇒ mx^{2} = 32 m^{2}x + 32

⇒ mx^{2} – 32 m^{2}x – 32 = 0 ……..(ii)

Line (i) touches the parabola x^{2} = 32y.

The quadratic equation (ii) in x has equal roots.

Discriminant = 0

⇒ (-32m^{2})^{2} – 4(m)(-32) = 0

⇒ 1024 m^{4} + 128m = 0

⇒ 128m (8m^{3} + 1) = 0

⇒ 8m^{3} + 1 = 0 …..[∵ m ≠ 0]

⇒ m^{3} = \(-\frac{1}{8}\)

⇒ m = \(-\frac{1}{2}\)

Substituting m = \(-\frac{1}{2}\) in (i), we get

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

⇒ \(y=-\frac{1}{2} x-2\)

⇒ x + 2y + 4 = 0, which is the equation of the common tangent.

Question 18.

Find the equation of the locus of a point, the tangents from which to the parabola y^{2} = 18x are such that sum of their slopes is -3.

Solution:

Given equation of the parabola is y^{2} = 18x

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

⇒ 4a = 18

⇒ a = \(\frac{9}{2}\)

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

⇒ y = mx + \(\frac{a}{m}\)

⇒ y = mx + \(\frac{9}{2m}\)

⇒ 2ym = 2xm^{2} + 9

⇒ 2xm^{2} – 2ym + 9 = 0

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

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

But, m_{1} + m_{2} = -3

\(\frac{y}{x}\) = -3

y = -3x, which is the required equation of locus.

Question 19.

The towers of a bridge, hung in the form of a parabola, have their tops 30 metres above the roadway and are 200 metres apart. If the cable is 5 metres above the roadway at the centre of the bridge, find the length of the vertical supporting cable 30 metres from the centre.

Solution:

Let CAB be the cable of the bridge and X’OX be the roadway.

Let A be the centre of the bridge.

From the figure, vertex of parabola is at A(0, 5).

Let the equation of parabola be

x^{2} = 4b(y – 5) …..(i)

Since the parabola passes through (100, 30).

Substituting x = 100 and y = 30 in (i), we get

⇒ 100^{2} = 4b (30 – 5)

⇒ 100^{2} = 4b(25)

⇒ 100^{2} = 100b

⇒ b = 100

Substituting the value of b in (i), we get

x^{2} = 400(y – 5) …..(ii)

Let l metres be the length of vertical supporting cable.

Then P(30, l) lies on (ii).

⇒ 30^{2} = 400(l – 5)

⇒ 900 = 400(l – 5)

⇒ \(\frac{9}{4}\) = l – 5

⇒ l = \(\frac{9}{4}\) + 5

⇒ l = \(\frac{9}{4}\) m = 7.25 m

The length of the vertical supporting cable is 7.25 m.

Question 20.

A circle whose centre is (4, -1) passes through the focus of the parabola x^{2} + 16y = 0. Show that the circle touches the directrix of the parabola.

Solution:

Given equation of the parabola is x^{2} + 16y = 0.

⇒ x^{2} = -16y

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

⇒ 4b = 16

⇒ b = 4

Focus = S(0, -b) = (0, -4)

Centre of the circle is C(4, -1) and it passes through focus S of the parabola.

Radius = CS

= \(\sqrt{(4-0)^{2}+(-1+4)^{2}}\)

= \(\sqrt{16+9}\)

= 5

Equation of the directrix is y – b = 0, i.e.,y – 4 = 0

Length of the perpendicular from centre C(4, -1) to the directrix

= \(\left|\frac{0(4)+1(-1)-4}{\sqrt{(0)^{2}+(1)^{2}}}\right|\)

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

= 5

= radius

∴ The circle touches the directrix of the parabola.