11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.2 Questions 12
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Question (1)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x

Solution

y2 = 12x
comparing to standard form y2 = 4ax we get a = 3
since equation of parabola is y2 = 12x, it is along x-axis
∴ focus f = (a, 0) = (3, 0)
Axis of parabola is x-axis
Equation of directrix is x = - a ⇒ x = -3
x + 3 = 0
Length of latus rectum = 4|a| = 12

Question (2)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = 6y

Solution

x2 = 6y
As equation of parabola is x2 = 6y, it is along y-axis,
Comparing to standard equation x2 = 4by
$b = \frac{6}{4} = \frac{3}{2}$
focus = (0, b) = $0,\frac{3}{2}$
axis is y-axis
Equation of directrix, y = -b $\Rightarrow y = \frac{{ - 3}}{2}$
2y + 3 = 0
Length of latus rectum = 4|b|
$ = 4\left( {\frac{3}{2}} \right) = 6$

Question (3)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x

Solution

Equation y2 = – 8x is along x-axis
comparing to standard form y2 = 4ax
a = -2
Focus = (a, 0) = (-2, 0)
axis of parabola is x-axis
Equation of directrix is x = -a ⇒ x + 2 = 0
LLR = 4|a| = 4|-2| = 8

Question (4)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y

Solution

x2 = – 16y
AS equation of parabola is x2 = -16y it is along y-axis comparing to standard equation x2 = 4aby we get , b = -4
focus = (0, b) = (0, -4)
Axis of parabola is y-axis
Equation of directrix is y=-b ⇒ y = 4
LLRE = 4|b| = 4}-4| = 16

Question (5)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 10x

Solution

Equation of parabola is y2 = 10x it is along x-axis
Comparing to std. equation y2 = 4ax
$a = \frac{{10}}{4} = \frac{5}{2}$
focus = (a, 0) $ = \left( {\frac{5}{2},0} \right)$
Equation of directrix is x = -a
$x = \frac{{ - 5}}{2} \Rightarrow 2x + 5 = 0$
LLR = 4|a| $ = 4\left| {\frac{5}{2}} \right| = 10$

Question (6)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = –9y

Solution

As equation is x2 = -9y
It is along y-axis
Comparing to standard equation x2 = 4by, $b = \frac{{ - 9}}{4}$
focus = (0, b) = (0, -9/4)
Axis = y-axis
Equation of directrix is y = -b
y = 9/4
LLR = 4|b| $ = 4\left| {\frac{{ - 9}}{4}} \right| = 9$

Question (7)

Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6

Solution

F(6, 0) directrix x = -6
As focus (6, 0) it is along x-axis a = 6
Equation of parabola along x-axis is y2 = 4ax
y2 = 4(6)x = 24x
y2 = 24x

Question (8)

Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

Solution

Focus (0, -3) directrix y = 3
x-coordinate of focus is 0
⇒ parabola is along x-axiis
F(0, b) = (0, -3)
⇒ b = -3
Equation of parabola along x-axis is
x2 = 4by
x2 = 4(-3)y
x2 = -12y

Question (9)

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

Solution

Vertex (0, 0) focus (3, 0) as y-coordinate of focus is zero, it is along x-axis
focus (a, 0) = (3, 0) ⇒ a = 3
Equation of parabola along x-axis x-axis is
y2 = 4ax = 4(3)x = 12x

Question (10)

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Solution

Vertex (0, 0), focus (–2, 0)
As y-coordinate of focus is zero, it is along x-axis
f(a, 0) = (-2, 0) ⇒ a = -2
Equation of parabola along x-axis is
y2 = 4ax
y2 = 4(-2)x
y2 = -8x

Question (11)

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis

Solution

Vertex (0, 0), passing through (5, 2) and axis is along x-axis
The axis of parabola is along x-axis
so the equation of its is y2 = 4ax ---(1)
It passes through (2, 3)
∴ 32 = 4a(2)
9 = 8a
$a = \frac{9}{8}$
Replace value of a in (1) we get
${y^2} = 4\left( {\frac{9}{8}} \right)x$
${y^2} = \left( {\frac{9}{2}} \right)x$
2y2 = 9x

Question (12)

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis

Solution

Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis
AS parabola is symmetric about y-axis, it is along y-axis, so it equation will be
x2 = 4by
It passes through (5, 2)
∴ 52 = 4b(2)
$b = \frac{{25}}{8}$
Replacing value of b we get
${x^2} = 4\left( {\frac{{25}}{8}} \right)y$
${x^2} = \left( {\frac{{25}}{2}} \right)y$
2x2 = 25y
Exercise 11.1⇐
⇒Exercise11.3