11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.2 Questions 12

Hi

Question (1)

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for ySolution

ycomparing to standard form y

since equation of parabola is y

∴ 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 xSolution

xAs equation of parabola is x

Comparing to standard equation x

$b = \frac{6}{4} = \frac{3}{2}$

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 ySolution

Equation ycomparing to standard form y

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 xSolution

xAS equation of parabola is x

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 ySolution

Equation of parabola is yComparing to std. equation y

$a = \frac{{10}}{4} = \frac{5}{2}$

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 xSolution

As equation is xIt is along y-axis

Comparing to standard equation x

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 = –6Solution

F(6, 0) directrix x = -6As focus (6, 0) it is along x-axis a = 6

Equation of parabola along x-axis is y

y

y

Question (8)

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

Focus (0, -3) directrix y = 3x-coordinate of focus is 0

⇒ parabola is along x-axiis

F(0, b) = (0, -3)

⇒ b = -3

Equation of parabola along x-axis is

x

x

x

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-axisfocus (a, 0) = (3, 0) ⇒ a = 3

Equation of parabola along x-axis x-axis is

y

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

y

y

y

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-axisSolution

Vertex (0, 0), passing through (5, 2) and axis is along x-axisThe axis of parabola is along x-axis

so the equation of its is y

It passes through (2, 3)

∴ 3

9 = 8a

$a = \frac{9}{8}$

Replace value of a in (1) we get

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

${y^2} = \left( {\frac{9}{2}} \right)x$

2y

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-axisSolution

Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axisAS parabola is symmetric about y-axis, it is along y-axis, so it equation will be

x

It passes through (5, 2)

∴ 5

$b = \frac{{25}}{8}$

Replacing value of b we get

${x^2} = 4\left( {\frac{{25}}{8}} \right)y$

${x^2} = \left( {\frac{{25}}{2}} \right)y$

2x