Maharashtra State Board Class 11 (FYJC / HSC) Mathematics & Statistics Solutions Chapter 7 Conic Sections Miscellaneous Exercise 7 - Digest Answers | Maharashtra Board Solutions

Q: Question 7: The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the endpoints of its latus rectum is ________ (A) 22 sq. units (B) 20 sq. units (C) 18 s

(C) 18 sq. units Hint: x2 = 12y 4b = 12 b = 3 Area of triangle = × AB × OS = × 4a × a = × 12 × 3 = 18 sq. units

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

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

Question 1 Maharashtra Board Solution

The line y = mx + 1 is a tangent to the parabola y2 = 4x, if m is ________ (A) 1 (B) 2 (C) 3 (D) 4

Solution & Step-by-Step Answer:

(A) 1 Hint: y2 = 4x Compare with y2 = 4ax ∴ a = 1 Equation of tangent is y = mx + 1 Compare with y = mx + = 1 ∴ a = m = 1

Question 2 Maharashtra Board Solution

The length of latus rectum of the parabola x2 – 4x – 8y + 12 = 0 is ________ (A) 4 (B) 6 (C) 8 (D) 10

Solution & Step-by-Step Answer:

(C) 8 Hint: Given equation of parabola is x2 – 4x – 8y + 12 = 0 ⇒ x2 – 4x = 8y – 12 ⇒ x2 – 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 Maharashtra Board Solution

If the focus of the parabola is (0, -3), its directrix is y = 3, then its equation is ________ (A) x2 = -12y (B) x2 = 12y (C) y2 = 12x (D) y2 = -12x

Solution & Step-by-Step Answer:

(A) x2 = -12y Hint: SP2 = PM2 ⇒ (x – 0)2 + (y + 3)2 = ⇒ x2 + y2 + 6y + 9 = y2 – 6y + 9 ⇒ x2 = -12y

Question 4 Maharashtra Board Solution

The co-ordinates of a point on the parabola y2 = 8x whose focal distance is 4 are ________ (A) (, ±2) (B) (1, ±2√2) (C) (2, ±4) (D) none of these

Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution

The end points of latus rectum of the parabola y2 = 24x are ________ (A) (6, ±12) (B) (12, ±6) (C) (6, ±6) (D) none of these

Solution & Step-by-Step Answer:

(A) (6, ±12)

Question 6 Maharashtra Board Solution

Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________ (A) y2 = 8x (B) y2 = 32x (C) y2 = 16x (D) x2 = 32y

Solution & Step-by-Step Answer:

(B) y2 = 32x Hint: Since directrix is parallel to Y-axis, The X-axis is the axis of the parabola. Let the equation of parabola be y2 = 4ax. Equation of directrix is x + 8 = 0 ∴ a = 8 ∴ required equation of parabola is y2 = 32x

Question 7 Maharashtra Board Solution

The area of the triangle formed by the lines joining the vertex of the parabola x2 = 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

Solution & Step-by-Step Answer:

Question 8 Maharashtra Board Solution

If P() is any point on the ellipse 9x2 + 25y2 = 225, S and S’ are its foci, then SP. S’P = ________ (A) 13 (B) 14 (C) 17 (D) 19

Solution & Step-by-Step Answer:

(C) 17 Hint: 9x2 + 25y2 = 225 Here, a = 5, b = 3 Eccentricity (e) = ∴ Coordinates of foci are S(4, 0) and S'(-4, 0) P(θ) = (a cos θ, b sin θ)

Question 9 Maharashtra Board Solution

The equation of the parabola having (2, 4) and (2, -4) as end points of its latus rectum is ________ (A) y2 = 4x (B) y2 = 8x (C) y2 = -16x (D) x2 = 8y

Solution & Step-by-Step Answer:

(B) y2 = 8x Hint: The given points lie in the 1st and 4th quadrants. ∴ Equation of the parabola is y2 = 4ax End points of latus rectum are (a, 2a) and (a, -2a) ∴ a = 2 ∴ required equation of parabola is y = 8x

Question 10 Maharashtra Board Solution

If the parabola y2 = 4ax passes through (3, 2), then the length of its latus rectum is ________ (A) (B) (C) (D) 4

Solution & Step-by-Step Answer:

(B) Hint: Length of latus rectum = 4a The given parabola passes through (3, 2) ∴ (2)2 = 4a(3) ∴ 4a =

Question 11 Maharashtra Board Solution

The eccentricity of rectangular hyperbola is (A) (B) (C) (D)

Solution & Step-by-Step Answer:

(C)

Question 12 Maharashtra Board Solution

The equation of the ellipse having one of the foci at (4, 0) and eccentricity is (A) 9x2 + 16y2 = 144 (B) 144x2 + 9y2 = 1296 (C) 128x2 + 144y2 = 18432 (D) 144x2 + 128y2 = 18432

Solution & Step-by-Step Answer:

Question 13 Maharashtra Board Solution

The equation of the ellipse having eccentricity and passing through (-8, 3) is (A) 4x2 + y2 = 4 (B) x2 + 4y2 = 100 (C) 4x2 + y2 = 100 (D) x2 + 4y2 = 4

Solution & Step-by-Step Answer:

(B) x2 + 4y2 = 100

Question 14 Maharashtra Board Solution

If the line 4x – 3y + k = 0 touches the ellipse 5x2 + 9y2 = 45, then the value of k is (A) 21 (B) ±3√21 (C) 3 (D) 3(21)

Solution & Step-by-Step Answer:

(B) ±3√21

Question 15 Maharashtra Board Solution

The equation of the ellipse is 16x2 + 25y2 = 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

Solution & Step-by-Step Answer:

(B) y = ±4

Question 16 Maharashtra Board Solution

The equation of the tangent to the ellipse 4x2 + 9y2 = 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

Solution & Step-by-Step Answer:

Question 17 Maharashtra Board Solution

Eccentricity of the hyperbola 16x2 – 3y2 – 32x – 12y – 44 = 0 is (A) (B) (C) (D)

Solution & Step-by-Step Answer:

(B) Hint: 16x2 – 3y2 – 32x – 12y – 44 = 0 ⇒ 16(x – 1)2 – 3(y + 2)2 = 48 ⇒ Here, a2 = 3 and b2 = 16

Question 18 Maharashtra Board Solution

Centre of the ellipse 9x2 + 5y2 – 36x – 50y – 164 = 0 is at (A) (2, 5) (B) (1, -2) (C) (-2, 1) (D) (0, 0)

Solution & Step-by-Step Answer:

(A) (2, 5) Hint: 9x2 + 5y2 – 36x – 50y – 164 = 0 ⇒ 9(x – 2)2 + 5(y – 5)2 = 325 ⇒ ⇒ centre of the ellipse = (2, 5)

Question 19 Maharashtra Board Solution

If the line 2x – y = 4 touches the hyperbola 4x2 – 3y2 = 24, the point of contact is (A) (1, 2) (B) (2, 3) (C) (3, 2) (D) (-2, -3)

Solution & Step-by-Step Answer:

Question 20 Maharashtra Board Solution

The foci of hyperbola 4x2 – 9y2 – 36 = 0 are (A) (±√13, 0) (B) (±√11, 0) (C) (±√12, 0) (D) (0, ±√12)

Solution & Step-by-Step Answer:

(A) (±√13, 0)

II. Answer the following.

Question 1 Maharashtra Board Solution

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 2y2 = 17x (ii) 5x2 = 24y

Solution & Step-by-Step Answer:

(i) Given equation of the parabola is 2y2 = 17x y2 = x Comparing this equation with y2 = 4ax, we get 4a = a = Co-ordinates of focus are S(a, 0), i.e., S(, 0) Equation of the directrix is x + a = 0 x + = 0 8x + 17 = 0 Length of latus rectum = 4a = 4() = Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a) i.e., and

(ii) Given equation of the parabola is 5x²= 24y
x²=
Comparing this equation with x²= 4by, we get
4b =
b =
Co-ordinates of focus are S(0, b), i.e., S(0, )
Equation of the directrix is y + b = 0
y + = 0
5y + 6 = 0
Length of latus rectum = 4b = 4() =
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b), i.e., and

Question 2 Maharashtra Board Solution

Find the cartesian co-ordinates of the points on the parabola y2 = 12x whose parameters are (i) 2 (ii) -3

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 12x Comparing this equation with y2 = 4ax, we get 4a = 12 ∴ a = 3 If t is the parameter of the point P on the parabola, then P(t) = (at2, 2at) i.e., x = at2 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)²and y = 2(3)(-3)
∴ x = 27 and y = -18
∴ The cartesian co-ordinates of the point on the parabola are (27, -18).

Question 3 Maharashtra Board Solution

Find the co-ordinates of a point of the parabola y2 = 8x having focal distance 10.

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 8x Comparing this equation with y2 = 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 y2 = 8x, we get y2 = 8(8) ∴ y = ±8 ∴ The co-ordinates of the points on the parabola are (8, 8) and (8, -8).

Question 4 Maharashtra Board Solution

Find the equation of the tangent to the parabola y2 = 9x at the point (4, -6) on it.

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 9x Comparing this equation with y2 = 4ax, we get 4a = 9 ∴ a = Equation of the tangent y2 = 4ax at (x1, y1) is yy1 = 2a(x + x1) The equation of the tangent at (4, -6) is y(-6) = 2()(x + 4) ⇒ -6y = (x + 4) ⇒ -12y = 9x + 36 ⇒ 9x + 12y + 36 = 0 ⇒ 3x + 4y + 12 = 0

Question 5 Maharashtra Board Solution

Find the equation of the tangent to the parabola y2 = 8x at t = 1 on it.

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 8x Comparing this equation with y2 = 4ax, we get 4a = 8 a = 2 t = 1 Equation of tangent with parameter t is yt = x + at2 ∴ The equation of tangent with t = 1 is y(1) = x + 2(1)2 y = x + 2 ∴ x – y + 2 = 0

Question 6 Maharashtra Board Solution

Find the equations of the tangents to the parabola y2 = 9x through the point (4, 10).

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 9x Comparing this equation with y2 = 4ax, we get 4a = 9 ∴ a = Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + y = mx + But, (4, 10) lies on the tangent. 10 = 4m + ⇒ 40m = 16m2+ 9 ⇒ 16m2 – 40m + 9 = 0 ⇒ 16m2 – 36m – 4m + 9 = 0 ⇒ 4m(4m – 9) – 1(4m – 9) = 0 ⇒ (4m – 9) (4m – 1) = 0 ⇒ 4m – 9 = 0 or 4m – 1 = 0 ⇒ m = or m = These are the slopes of the required tangents. By slope point form, y – y1 = m(x – x1), the equations of the tangents are y – 10 = (x – 4) or y – 10 = (x – 4) ⇒ 4y – 40 = 9x – 36 or 4y – 40 = x – 4 ⇒ 9x – 4y + 4 = 0 or x – 4y + 36 = 0

Question 7 Maharashtra Board Solution

Show that the two tangents drawn to the parabola y2 = 24x from the point (-6, 9) are at the right angle.

Solution & Step-by-Step Answer:

Given the equation of the parabola is y2 = 24x. Comparing this equation with y2 = 4ax, we get 4a = 24 ⇒ a = 6 Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + ⇒ y = mx + But, (-6, 9) lies on the tangent 9 = -6m + ⇒ 9m = -6m2 + 6 ⇒ 6m2 + 9m – 6 = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1m2 = -1 Tangents drawn to the parabola y2 = 24x from the point (-6, 9) are at a right angle.

Alternate method:
Comparing the given equation with y²= 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 y²= 24x from the point (-6, 9) are at the right angle.

Question 8 Maharashtra Board Solution

Find the equation of the tangent to the parabola y2 = 8x which is parallel to the line 2x + 2y + 5 = 0. Find its point of contact.

Solution & Step-by-Step Answer:

Given the equation of the parabola is y2 = 8x. Comparing this equation with y2 = 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 y2 = 4ax having slope m is y = mx + Equation of the tangent is y = -x + x + y + 2 = 0 Point of contact = = = (2, -4)

Question 9 Maharashtra Board Solution

A line touches the circle x2 + y2 = 2 and the parabola y2 = 8x. Show that its equation is y = ±(x + 2).

Solution & Step-by-Step Answer:

Given equation of the parabola is y2 = 8x Comparing this equation with y2 = 4ax, we get 4a = 8 a = 2 Equation of tangent to given parabola with slope m is y = mx + m2x – my + 2 = 0 ….(i) Equation of the circle is x2 + y2 = 2 Its centre = (0, 0) and Radius = √2 Line (i) touches the circle. Length of perpendicular from the centre to the line (i) = radius ⇒ = √2 ⇒ = 2 ⇒ m4 + m2 – 2 – 0 ⇒ (m2 + 2)(m2 – 1) = 0 Since m2 ≠ -2, m2 – 1 = 0 ⇒ m = ±1 When m = 1, equation of the tangent is y = (1)x + y = (x + 2) …..(i) When m = -1, equation of the tangent is y = (-1)x + 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 Maharashtra Board Solution

Two tangents to the parabola y2 = 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 y2 = 8(x + 2).

Solution & Step-by-Step Answer:

Given parabola is y2 = 8x Comparing with y2 = 4ax, we get, 4a = 8 ⇒ a = 2 Let M(t1) and N(t2) be any two points on the parabola. The equations of tangents at M and N are yt1 = x + …..(1) yt2 = x + …(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) ⇒ yt1 = ⇒ y = 2t1 …..(t1 ≠ 0 otherwise tangent at M will be x = 0) ⇒ P = (0, 2t1) Similarly, Q = (0, 2t2) It is given that PQ = 4 ∴ |2t1 – 2t2| = 4 ∴ |t1 – t2| = 2 …..(3) Let R = (x1, y1) 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(t1 – t2) = y(t1 – t2) = 2(t1 – t2)(t1 + t2) ∴ y = 2(t1 + t2) …..[∵ M, N are distinct ∴ t1 ≠ t2] i.e., y1 = 2(t1 + t2) …..(4) ∴ from (1), we get 2t1(t1 + t2) = x + ∴ 2t1t2 = x i.e. x1 = 2t1t2 …..(5) To find the equation of locus of R(x1, y1), we eliminate t1 and t2 from the equations (3), (4) and (5). We know that, (t1 + t2)2 = (t1 + t2)2 + 4t1t2 ⇒ …[By (3), (4) and (5)] ⇒ = 16 + 8x1 = 8(x1 + 2) Replacing x1 by x and y1 by y, the equation of required locus is y2 = 8(x + 2).

Question 11 Maharashtra Board Solution

The slopes of the tangents drawn from P to the parabola y2 = 4ax are m1 and m2, showing that (i) m1 – m2 = k (ii) = k, where k is a constant.

Solution & Step-by-Step Answer:

Let P(x1, y1) be any point on the parabola y2 = 4ax. Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + This tangent passes through P(x1, y1). y1 = mx1 + my1 = m2x1 + a m2x1 – my1 + a = 0 This is a quadratic equation in ‘m’. The roots m1 and m2 of this quadratic equation are the slopes of the tangents drawn from P. ∴ m1 + m2 = , m1m2 = Since (x1, y1) and a are constants, m1 – m2 is a constant. ∴ m1 – m2 = k, where k is constant.

(ii) Since (x₁, y₁) and a are constants, m₁m₂is a constant.
= k, where k is a constant.

Question 12 Maharashtra Board Solution

The tangent at point P on the parabola y2 = 4ax meets the Y-axis in Q. If S is the focus, show that SP subtends a right angle at Q.

Solution & Step-by-Step Answer:

Let P(, 2at1) be a point on the parabola and S(a, 0) be the focus of parabola y2 = 4ax Since the tangent passing through point P meet Y-axis at point Q, equation of tangent at P(, 2at1) is yt1 = x + …..(i) ∴ Point Q lie on tangent ∴ put x = 0 in equation (i) yt1 = y = at1 ∴ Co-ordinate of point Q(0, at1) S = (a, 0), P(, 2at1), Q(0, at1) ∴ SP subtends a right angle at Q.

Question 13 Maharashtra Board Solution

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) (b) 16x2 + 25y2 = 400 (c) (d) x2 – y2 = 16

Solution & Step-by-Step Answer:

(a) Given equation of the ellipse is Comparing this equation with , we get a2 = 25 and b2 = 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 =
∴ e = =
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)
i.e., S(5(), 0) and S'(-5(), 0),
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±
i.e., x = ±
i.e., x = ±

(iv) Length of latus rectum =

(v) Distance between foci = 2ae = 2 (5) () = 8

(vi) Distance between directrices = = =

(b) Given equation of the ellipse is 16x²+ 25y²= 400

Comparing this equation with , we get
a²= 25 and b²= 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) b²= a²(1 – e²)
16 = 25(1 – e²)
= 1 – e²
e²= 1 –
e²=
e = ……[∵ 0 < e < 1]
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(), 0) and S'(-5(), 0),
i.e., S(3, 0) and S'(-3, 0)

(iii) Equations of the directrices are x = ±
i.e., x = ±
i.e., x = ±

(iv) Length of latus rectum =

(v) Distance between foci = 2ae = 2(5)() = 6

(vi) Distance between directrices =

(c) Given equation of the hyperbola
Comparing this equation with
a²= 144 and b²= 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) b²= a²(e²– 1)
25 = 144 (e²– 1)
= e²– 1
e²= 1 +
e²=
e = …….[∵ e > 1]
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0)
i.e., S(12(), 0) and S'(-12(), 0)
i.e., S(13, 0) and S'(-13, 0)

(iii) Equations of the directrices are x =
i.e., x =
i.e., x =

(iv) Length of latus rectum = =

(v) Distance between foci = 2ae = 2(12)() = 26

(vi) Distance between directrices = =

(d) Given equation of the hyperbola is x²– y²= 16
∴
Comparing this equation with , we get
a²= 16 and b²= 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 = ±
∴x = ±
∴ x = ±2√2

(iv) Length of latus rectum = = = 8

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

(vi) Distance between directrices = = = 4√2.

Question 14 Maharashtra Board Solution

Find the equation of the ellipse in standard form if (i) eccentricity = 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 & Step-by-Step Answer:

(i) Let the required equation of ellipse be , where a > b. Given, eccentricity (e) = Distance between foci = 2ae Given, distance between foci = 6 ∴ 2ae = 6 ∴ 2a() = 6 ∴ = 6 ∴ a = 8 ∴ a2 = 64 Now, b2 = a2 (1 – e2) = = = 64() = 55 ∴ The required equation of the ellipse is

(ii) Let the equation of the ellipse be
……(1)
Then length of major axis = 2a = 10
∴ a = 5
Also, distance between foci= 2ae = 8
∴ 2 × 5 × e = 8
∴ e =
∴ b²= a²(1 – e²)
= 25(1 – )
= 9
∴ from (1), the equation of the required ellipse is

(iii) Let the required equation of ellipse be , 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

∴ …..(i)
Substituting x = 2 and y = -2 in equation of ellipse, we get

∴ ……(ii)
Let = A and = 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 =
Substituting A = in (iv), we get
4() + 4B = 1
∴ + 4B = 1
∴ 4B = 1 –
∴ 4B =
∴ B =
Since = A and = B
and
∴ a²= and b²=
∴ The required equation of ellipse is

i.e., 3x²+ 5y²= 32.

Question 15 Maharashtra Board Solution

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

Solution & Step-by-Step Answer:

Let the equation of the ellipse be It is given that, distance between directrices is three times the distance between the foci. ∴ = 3(2ae) ∴ 1 = 3e2 ∴ e2 = ∴ e = …..[∵ 0 < e < 1]

Question 16 Maharashtra Board Solution

For the hyperbola , prove that SA. S’A = 25, where S and S’ are the foci and A is the vertex.

Solution & Step-by-Step Answer:

Given equation of the hyperbola is Comparing this equation with , we get a2 = 100 and b2 = 25 ∴ a = 10 and b = 5 ∴ Co-ordinates of vertex is A(a, 0), i.e., A(10, 0) Eccentricity, e = = = = = Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0) i.e., S(10(), 0) and S'(-10(), 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 Maharashtra Board Solution

Find the equation of the tangent to the ellipse passing through the point (2, -2).

Solution & Step-by-Step Answer:

Given equation of the ellipse is Comparing this equation with , we get a2 = 5 and b2 = 4 Equations of tangents to the ellipse having slope m are y = mx ± Since (2, -2) lies on both the tangents, -2 = 2m ± ∴ -2 – 2m = ± Squaring both the sides, we get 4m2 + 8m + 4 = 5m2 + 4 ∴ m2 – 8m = 0 ∴ m(m – 8) = 0 ∴ m = 0 or m = 8 These are the slopes of the required tangents. ∴ By slope point form y – y1 = m(x – x1), 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 Maharashtra Board Solution

Find the equation of the tangent to the ellipse x2 + 4y2 = 100 at (8, 3).

Solution & Step-by-Step Answer:

Given equation of ellipse is x2 + 4y2 = 100 ∴ Comparing this equation with , we get a2 = 100 and b2 = 25 Equation of tangent to the ellipse at (x1, y1) is Equation of tangent at (8, 3) is 2x + 3y = 25

Question 19 Maharashtra Board Solution

Show that the line 8y + x = 17 touches the ellipse x2 + 4y2 = 17. Find the point of contact.

Solution & Step-by-Step Answer:

Question 20 Maharashtra Board Solution

Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.

Solution & Step-by-Step Answer:

Given equation of the ellipse is 4x2 + 5y2 = 20. ∴ Comparing this equation with , we get a2 = 5 and b2 = 4 Since inclinations of tangents are θ1 and θ2, m1 = tan θ1 and m2 = tan θ2 Equation of tangents to the ellipse having slope m are y = mx ± ∴ y = mx ± ∴ y – mx = ± Squaring both the sides, we get y2 – 2mxy + m2x2 = 5m2 + 4 ∴ (x2 – 5)m2 – 2xym + (y2 – 4) = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. ∴ m1 + m2 = Given, tan θ1 + tan θ2 = 2 ∴ m1 + m2 = 2 ∴ ∴ xy = x2 – 5 ∴ x2 – xy – 5 = 0, which is the required equation of the locus of P.

Question 21 Maharashtra Board Solution

Show that the product of the lengths of its perpendicular segments drawn from the foci to any tangent line to the ellipse is equal to 16.

Solution & Step-by-Step Answer:

Given equation of the ellipse is Comparing this equation with , we get ∴ a2 = 25, b2 = 16 ∴ a = 5, b = 4 We know that e = ∴ e = = ae = 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 having slope m are y = mx ± Equation of one of the tangents to the ellipse is y = mx + ∴ mx – y + = 0 …..(i) p1 = length of perpendicular segment from S(3, 0) to the tangent (i) p2 = length of perpendicular segment from S'(-3, 0) to the tangent (i)

Question 22 Maharashtra Board Solution

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 and distance between foci is 12. (iii) length of the conjugate axis is 3 and the distance between the foci is 5.

Solution & Step-by-Step Answer:

(i) Let the required equation of hyperbola be Length of conjugate axis = 2b Given, length of conjugate axis = 5 2b = 5 b = b2 = Distance between foci = 2ae Given, distance between foci = 13 2ae = 13 ae = a2e2 = Now, b2 = a2(e2 – 1) b2 = a2e2 – a2 = – a2 a2 = = 36 ∴ The required equation of hyperbola is i.e.,

(ii) Let the required equation of hyperbola be
Given, eccentricity (e) =
Distance between foci = 2ae
Given, distance between foci = 12
∴ 2ae = 12
∴ 2a() = 12
∴ 3a = 12
∴ a = 4
∴ a²= 16
Now, b²= a²(e²– 1)
∴ b²=
∴ b²= 16( – 1)
∴ b²= 16()
∴ b²= 20
∴ The required equation of hyperbola is

(iii) Let the required equation of hyperbola be
Length of conjugate axis = 2b
Given, length of conjugate axis = 3
∴ 2b = 3
∴ b =
∴ b²=
Distance between foci = 2ae
Given, distance between foci = 5
∴ 2ae = 5
∴ ae =
∴ a²e²=
Now, b²= a²(e²– 1)
∴ b²= a²e²– a²
∴ = – a²
∴ a²=
∴ a²= 4
∴ The required equation of hyperbola is
i.e.,

Question 23 Maharashtra Board Solution

Find the equation of the tangent to the hyperbola, (i) 7x2 – 3y2 = 51 at (-3, -2) (ii) x = 3 sec θ, y = 5 tan θ at θ = π/3 (iii) at P(30°).

Solution & Step-by-Step Answer:

(i) Given equation of the hyperbola is 7x2 – 3y2 = 51

(ii) Given, equation of the hyperbola is
x = 3 sec θ, y = 5 tan θ
Since sec²θ – tan²θ = 1,

Comparing this equation with , we get
a²= 9 and b²= 25
a = 3 and b = 5
Equation of tangent at P(θ) is

∴ Equation of tangent at P(π/3) is

10x – 3√3 y = 15

(iii) Given equation of hyperbola is
Comparing this equation with , we get
a²= 25 and b²= 16
a = 5 and b = 4
Equation of tangent at P(θ) is

The equation of tangent at P(30°) is

8x – 5y = 20√3

Question 24 Maharashtra Board Solution

Show that the line 2x – y = 4 touches the hyperbola 4x2 – 3y2 = 24. Find the point of contact.

Solution & Step-by-Step Answer:

Given equation of die hyperbola is 4x2 – 3y2 = 24. ∴ Comparing this equation with , we get a2 = 6 and b2 = 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 , we must have c2 = a2m2 – b2 c2 = (-4)2 = 16 a2m2 – b2 = 6(2)2 – 8 = 24 – 8 = 16 ∴ The given line is a tangent to the given hyperbola and point of contact = = = (3, 2)

Question 25 Maharashtra Board Solution

Find the equations of the tangents to the hyperbola 3x2 – y2 = 48 which are perpendicular to the line x + 2y – 7 = 0.

Solution & Step-by-Step Answer:

Given the equation of the hyperbola is 3x2 – y2 = 48. ∴ Comparing this equation with , we get a2 = 16 and b2 = 48 Slope of the line x + 2y – 7 = 0 is Since the given line is perpendicular to the tangents, slope of the required tangent (m) = 2 Equations of tangents to the ellipse having slope m are y = mx ± y = 2x ± y = 2x ± √16 ∴ y = 2x ± 4

Question 26 Maharashtra Board Solution

Two tangents to the hyperbola make angles θ1, θ2, with the transverse axis. Find the locus of their point of intersection if tan θ1 + tan θ2 = k.

Solution & Step-by-Step Answer:

Given equation of the hyperbola is Let θ1 and θ2 be the inclinations. m1 = tan θ1, m2 = tan θ2 Let P(x1, y1) be a point on the hyperbola Equation of a tangent with slope ‘m’ to the hyperbola is y = mx ± This tangent passes through P(x1, y1). y1 = mx1 ± (y1 – mx1)2 = a2m2 – b2 ……(i) This is a quadratic equation in ‘m’. It has two roots say m1 and m2, which are the slopes of two tangents drawn from P. ∴ m1 + m2 = Since tan θ1 + tan θ2 = k, ∴ P(x1, y1) moves on the curve whose equation is k(x2 – a2) = 2xy.

Chapter 7 Conic Sections Miscellaneous Exercise 7 Solutions