Maharashtra Board Text books

Maharashtra Board Class 11 Maths Part 1 Chapter 7 Conic Sections Ex 7.1 Solution

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:
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(ii) Given equation of the parabola is  = -20x.
Comparing this equation with  = -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).

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(iv) Given equation of the parabola is  = -8y.
Comparing this equation with  = -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).

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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:

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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:

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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  =-4(7)x, i.e.,  = -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  = 4ax or  = -4ax.
Since the parabola passes through (1, -6), it lies in the 4th quadrant.
Required parabola is  = 4ax.
Substituting x = 1 and y = -6 in  = 4ax, we get
⇒ (-6 = 4a(1)
⇒ 36 = 4a
⇒ a = 9
∴ The required equation of the parabola is  = 4(9)x, i.e.,  = 36x.

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Question 6.
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Maharashtra Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.1 Q6
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∴ The parameter of the given point is −92

Question 7.
Find the focal distance of a point on the parabola  = 16x whose ordinate is 2 times the abscissa.
Solution:

Given the equation of the parabola is  = 16x.
Comparing this equation with  = 4ax, we get
⇒ 4a = 16
⇒ a = 4
Since ordinate is 2 times the abscissa,
y = 2x
Substituting y = 2x in  = 16x, we get
⇒ (2x = 16x
⇒ 4 = 16x
⇒ 4 – 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.

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Question 9.
For the parabola  = 4x, find the coordinates of the point whose focal distance is 17.
Solution:

Given the equation of the parabola is  = 4x.
Comparing this equation with  = 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
⇒  = 4(16)
⇒  = 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  = 4ax passing through the point (2, -6).
Solution:

Given equation of the parabola is  = 4ax and it passes through point (2, -6).
Substituting x = 2 and y = -6 in  = 4ax, we get
⇒ (-6 = 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  = 12y to the endpoints of the latus rectum.
Solution:

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Given the equation of the parabola is  = 12y.
Comparing this equation with  = 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
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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:

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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  = 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
⇒ 1 = 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 =  – 2x + 3.
Solution:

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⇒ 4y – 7 = 0

Question 14.
Find the equation of tangent to the parabola
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Question 15.
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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.
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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, ) and Q be (0, ).
Then, from (i) and (ii), we get
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∴ Equation of locus of R is  = 8(x + 2).

Question 17.
Find the equation of common tangent to the parabolas  = 4x and  = 32y.
Solution:

Given equation of the parabola is  = 4x
Comparing this equation with  = 4ax, we get
⇒ 4a = 4
⇒ a = 1
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⇒ 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 = 18x are such that sum of their slopes is -3.
Solution:

Given equation of the parabola is  = 18x
Comparing this equation with = 4ax, we get
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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
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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  + 16y = 0. Show that the circle touches the directrix of the parabola.
Solution:

Given equation of the parabola is  + 16y = 0.
⇒  = -16y
Comparing this equation with = -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.
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= radius
∴ The circle touches the directrix of the parabola.