Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1
Question 1
Maharashtra Board Solution
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) 5y2 = 24x (ii) y2 = -20x (iii) 3x2 = 8y (iv) x2 = -8y (v) 3y2 = -16x
Solution & Step-by-Step Answer:
(i) Given equation of the parabola is 5y2 = 24x. ⇒ 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 ⇒ 5x + 6 = 0 Length of latus rectum = 4a = 4() = Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a), ⇒ and
(ii) Given equation of the parabola is y2= -20x.
Comparing this equation with y2= -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 3x2= 8y
⇒ x2= y
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
⇒ 3y + 2 = 0
Length of latus rectum = 4b = 4() =
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b),
⇒ and .
(iv) Given equation of the parabola is x2= -8y.
Comparing this equation with x2= -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 3y2= -16x.
⇒ y2= x
Comparing this equation withy = -4ax, we get
⇒ 4a =
⇒ a =
Co-ordinates of focus are S(-a, 0), i.e., (, 0)
Equation of the directrix is x – a = 0,
⇒ x – = 0
⇒ 3x – 4 = 0
Length of latus rectum = 4a = 4() =
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
i.e., and
Question 2
Maharashtra Board Solution
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 & Step-by-Step Answer:
Vertex of the parabola is at origin (0, 0) and its axis is along Y-axis. Equation of the parabola can be either x2 = 4by or x2 = -4by Since the parabola passes through (-10, -5), it lies in 3rd quadrant. Required parabola is x2 = -4by. Substituting x = -10 and y = -5 in x2 = -4by, we get ⇒ (-10)2 = -4b(-5) ⇒ b = = 5 ∴ The required equation of the parabola is x2 = -4(5)y, i.e., x2 = -20y.
Question 3
Maharashtra Board Solution
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 & Step-by-Step Answer:
Vertex of the parabola is at the origin (0, 0) and its axis is along X-axis. Equation of the parabola can be either y2 = 4ax or y2 = -4ax. Since the parabola passes through (3, 4), it lies in the 1st quadrant. Required parabola is y2 = 4ax. Substituting x = 3 and y = 4 in y2 = 4ax, we get ⇒ (4)2 = 4a(3) ⇒ a = The required equation of the parabola is y2 = 4()x ⇒ 3y2 = 16x
Question 4
Maharashtra Board Solution
Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).
Solution & Step-by-Step Answer:
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 y2 =-4(7)x, i.e., y2 = -28x.
Question 5
Maharashtra Board Solution
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 & Step-by-Step Answer:
(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis. Equation of the parabola can be either y2 = 4ax or y2 = -4ax. Since the parabola passes through (1, -6), it lies in the 4th quadrant. Required parabola is y2 = 4ax. Substituting x = 1 and y = -6 in y2 = 4ax, we get ⇒ (-6)2 = 4a(1) ⇒ 36 = 4a ⇒ a = 9 ∴ The required equation of the parabola is y2 = 4(9)x, i.e., y2 = 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 y2= 4ax or y2= -4ax.
Since the parabola passes through (2, 3), it lies in 1st quadrant.
∴ Required parabola is y2= 4ax.
Substituting x = 2 and y = 3 in y2= 4ax, we get
⇒ (3)2= 4a(2)
⇒ 9 = 8a
⇒ a =
The required equation of the parabola is
y2= 4()x
⇒ y2= x
⇒ 2y2= 9x.
Question 6
Maharashtra Board Solution
For the parabola 3y2 = 16x, find the parameter of the point: (i) (3, -4) (ii) (27, -12)
Solution & Step-by-Step Answer:
Given the equation of the parabola is 3y2 = 16x. ⇒ y2 = x Comparing this equation with y2 = 4ax, we get ⇒ 4a = ⇒ a = 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 point is (3, -4) Substituting x = 3, y = -4 and a = in (i), we get 3 = t2 and -4 = 2() t ∴ The parameter of the given point is
(ii) Given point is (27, -12)
Substituting x = 27, y = -12 and a = in (i), we get
∴ The parameter of the given point is
Question 7
Maharashtra Board Solution
Find the focal distance of a point on the parabola y2 = 16x whose ordinate is 2 times the abscissa.
Solution & Step-by-Step Answer:
Given the equation of the parabola is y2 = 16x. Comparing this equation with y2 = 4ax, we get ⇒ 4a = 16 ⇒ a = 4 Since ordinate is 2 times the abscissa, y = 2x Substituting y = 2x in y2 = 16x, we get ⇒ (2x)2 = 16x ⇒ 4x2 = 16x ⇒ 4x2 – 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
Maharashtra Board Solution
Find coordinates of the point on the parabola. Also, find focal distance. (i) y2 = 12x whose parameter is (ii) 2y2 = 7x whose parameter is -2
Solution & Step-by-Step Answer:
(i) 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) Given, t = Substituting a = 3 and t = in (i), we get x = 3()2 and y = 2(3)() x = and y = 2 The co-ordinates of the point on the parabola are (, 2) ∴ Focal distance = x + a = + 3 =
(ii) Given equation of the parabola is 2y2= 7x.
⇒ y2= x
Comparing this equation with y2= 4ax, we get
⇒ 4a =
⇒ a =
If t is the parameter of the point P on the parabola, then
P(t) = (at2, 2at)
i.e., x = at2and y = 2at …..(i)
Given, t = -2
Substituting a = and t = -2 in (i), we get
x = (-2)2 and y = 2()(-2)
x = and y =
The co-ordinates of the point on the parabola are (, )
∴ Focal distance = x + a
= +
=
Question 9
Maharashtra Board Solution
For the parabola y2 = 4x, find the coordinates of the point whose focal distance is 17.
Solution & Step-by-Step Answer:
Given the equation of the parabola is y2 = 4x. Comparing this equation with y2 = 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 y2 = 4x, we get ⇒ y2 = 4(16) ⇒ y2 = 64 ⇒ y = ±8 ∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).
Question 10
Maharashtra Board Solution
Find the length of the latus rectum of the parabola y2 = 4ax passing through the point (2, -6).
Solution & Step-by-Step Answer:
Given equation of the parabola is y2 = 4ax and it passes through point (2, -6). Substituting x = 2 and y = -6 in y2 = 4ax, we get ⇒ (-6)2 = 4a(2) ⇒ 4a = 18 ∴ Length of latus rectum = 4a = 18 units
Question 11
Maharashtra Board Solution
Find the area of the triangle formed by the line joining the vertex of the parabola x2 = 12y to the endpoints of the latus rectum.
Solution & Step-by-Step Answer:
Given the equation of the parabola is x2 = 12y. Comparing this equation with x2 = 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’ = × l(LL’) × l(OS) = × 12 × 3 Area of ∆OLL’ = 18 sq. units
Question 12
Maharashtra Board Solution
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution & Step-by-Step Answer:
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 y2 = 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 ⇒ 102 = 4a(5) ⇒ 100 = 20a ⇒ a = 5 Focus is at (a, 0), i.e., (5, 0)
Question 13
Maharashtra Board Solution
Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x2 – 2x + 3.
Solution & Step-by-Step Answer:
Given equation of the parabola is y = x2 – 2x + 3 ⇒ y = x2 – 2x + 1 + 2 ⇒ y – 2 = (x – 1)2 ⇒ (x – 1)2 = y – 2 Comparing this equation with X2 = 4bY, we get X = x – 1, Y = y – 2 ⇒ 4b = 1 ⇒ b = 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 = ⇒ x = 1 and y = The co-ordinates of focus are (1, ) Equation of the axis is X = 0 x – 1 = 0, i.e., x = 1 Equation of directrix is Y + b = 0 ⇒ y – 2 + = 0 ⇒ y – = 0 ⇒ 4y – 7 = 0
Question 14
Maharashtra Board Solution
Find the equation of tangent to the parabola (i) y2 = 12x from the point (2, 5) (ii) y2 = 36x from the point (2, 9)
Solution & Step-by-Step Answer:
(i) Given equation of the parabola is y2 = 12x. Comparing this equation with y2 = 4ax, we get ⇒ 4a = 12 ⇒ a = 3 Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + Since the tangent passes through the point (2, 5) ⇒ 5 = 2m + ⇒ 5m = 2m2 + 3 ⇒ 2m2 – 5m + 3 = 0 ⇒ 2m2 – 2m – 3m + 3 = 0 ⇒ 2m(m – 1) – 3(m – 1) = 0 ⇒ (m- 1)(2m – 3) = 0 ⇒ m = 1 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 – 5 = 1(x – 2) and y – 5 = (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 y2= 36x.
Comparing this equation with y2= 4ax, we get
⇒ 4a = 36
⇒ a = 9
Equation of tangent to the parabola y2= 4ax having slope m is
y = mx +
Since the tangent passes through the point (2, 9),
⇒ 9 = 2m +
⇒ 9m = 2m2+ 9
⇒ 2m2– 9m + 9 = 0
⇒ 2m2– 6m – 3m + 9 = 0
⇒ 2m(m – 3) – 3(m – 3) = 0
⇒ (m – 3)(2m – 3) = 0
⇒ m = 3 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 – 9 = 3(x – 2) and y – 9 = (x – 2)
⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6
⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0
Question 15
Maharashtra Board Solution
If the tangents drawn from the point (-6, 9) to the parabola y2 = kx are perpendicular to each other, find k.
Solution & Step-by-Step Answer:
Given equation of the parabola is y2 = kx Comparing this equation with y2 = 4ax, we get ⇒ 4a = k ⇒ a = Equation of tangent to the parabola y2 = 4ax having slope m is y = mx + Since the tangent passes through the point (-6, 9), ⇒ 9 = -6m + ⇒ 36m = -24m2 + k ⇒ 24m2 + 36m – k = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1m2 = Since the tangents are perpendicular to each other, m1m2 = -1 ⇒ = -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
Maharashtra Board Solution
Two tangents to the parabola y2 = 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 y2 = 8(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 at A(t1) is y t1 = x + 2 …….(i) Equation of tangent to given parabola at B(t2) is y t2 = x + 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, k1) and Q be (0, k2). Then, from (i) and (ii), we get ∴ Equation of locus of R is y2 = 8(x + 2).
Question 17
Maharashtra Board Solution
Find the equation of common tangent to the parabolas y2 = 4x and x2 = 32y.
Solution & Step-by-Step Answer:
Given equation of the parabola is y2 = 4x Comparing this equation with y2 = 4ax, we get ⇒ 4a = 4 ⇒ a = 1 Let the equation of common tangent be y = mx + …..(i) Substituting y = mx + in x2 = 32y, we get ⇒ x2 = 32(mx + ) = 32 mx + ⇒ mx2 = 32 m2x + 32 ⇒ mx2 – 32 m2x – 32 = 0 ……..(ii) Line (i) touches the parabola x2 = 32y. The quadratic equation (ii) in x has equal roots. Discriminant = 0 ⇒ (-32m2)2 – 4(m)(-32) = 0 ⇒ 1024 m4 + 128m = 0 ⇒ 128m (8m3 + 1) = 0 ⇒ 8m3 + 1 = 0 …..[∵ m ≠ 0] ⇒ m3 = ⇒ m = Substituting m = in (i), we get ⇒ ⇒ ⇒ x + 2y + 4 = 0, which is the equation of the common tangent.
Question 18
Maharashtra Board Solution
Find the equation of the locus of a point, the tangents from which to the parabola y2 = 18x are such that sum of their slopes is -3.
Solution & Step-by-Step Answer:
Given equation of the parabola is y2 = 18x Comparing this equation with y2 = 4ax, we get ⇒ 4a = 18 ⇒ a = Equation of tangent to the parabola y2 = 4ax having slope m is ⇒ y = mx + ⇒ y = mx + ⇒ 2ym = 2xm2 + 9 ⇒ 2xm2 – 2ym + 9 = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1 + m2 = But, m1 + m2 = -3 = -3 y = -3x, which is the required equation of locus.
Question 19
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 x2 = 4b(y – 5) …..(i) Since the parabola passes through (100, 30). Substituting x = 100 and y = 30 in (i), we get ⇒ 1002 = 4b (30 – 5) ⇒ 1002 = 4b(25) ⇒ 1002 = 100b ⇒ b = 100 Substituting the value of b in (i), we get x2 = 400(y – 5) …..(ii) Let l metres be the length of vertical supporting cable. Then P(30, l) lies on (ii). ⇒ 302 = 400(l – 5) ⇒ 900 = 400(l – 5) ⇒ = l – 5 ⇒ l = + 5 ⇒ l = m = 7.25 m The length of the vertical supporting cable is 7.25 m.
Question 20
Maharashtra Board Solution
A circle whose centre is (4, -1) passes through the focus of the parabola x2 + 16y = 0. Show that the circle touches the directrix of the parabola.
Solution & Step-by-Step Answer:
Given equation of the parabola is x2 + 16y = 0. ⇒ x2 = -16y Comparing this equation with x2 = -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 = = = 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 = = = 5 = radius ∴ The circle touches the directrix of the parabola.