Exercise 10.2 (Revised) - Chapter 11 - Conic Sections - Ncert Solutions class 11 - Maths
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Chapter 10: Conic Sections - NCERT Solutions for Class 11 Maths
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
Ex 10.2 Question 1.
$y^2=12 x$
Answer.
Given: Equation of parabola
$
y^2=12 x
$
Comparing with $y^2=4 a x$, we have $4 a=12$
$
\Rightarrow a=3
$
$\therefore$ Coordinates of focus are $(3,0)$.
Axis of parabola is $x$ - axis or $y=0$
Equation of the directrix is $x=-3$
$
\Rightarrow x+3=0
$
Length of latus rectum $=4 \times 3=12$ units
Ex 10.2 Question 2.
$x^2=6 y$
Answer.
Given: Equation of parabola
$
x^2=6 y
$
Comparing with $x^2=4 a y$, we have $4 a=6$
$
\Rightarrow a=\frac{3}{2}
$
$\therefore$ Coordinates of focus are $\left(0, \frac{3}{2}\right)$.
Axis of parabola is $y$ axis or $x=0$
Equation of the directrix is $y=\frac{-3}{2}$
$
\Rightarrow 2 y+3=0
$
Length of latus rectum $=\frac{4 \times 3}{2}=6$ units
Ex 10.2 Question 3.
$y^2=-8 x$
Answer.
Given: Equation of parabola
$
y^2=-8 x
$
Comparing with $y^2=-4 a x$, we have $4 a=8$
$
\Rightarrow a=2
$
$\therefore$ Coordinates of focus are $(-2,0)$.
Axis of parabola is $\mathrm{x}$ axis or $y=0$
Equation of the directrix is $x=2$
$
\Rightarrow x-2=0
$
Length of latus rectum $=4 \times 2=8$ units
Ex 10.2 Question 4.
$x^2=-16 y$
Answer.
Given: Equation of parabola
$
x^2=-16 y
$
Comparing with $x^2=-4 a y$, we have $4 a=16$
$
\Rightarrow a=4
$
$\therefore$ Coordinates of focus are $(0,-4)$.
Axis of parabola is $y$ axis or $x=0$
Equation of the directrix is $y=4$
$
\Rightarrow y-4=0
$
Length of latus rectum $=4 \times 4=16$ units
Ex 10.2 Question 5.
$y^2=10 x$
Answer.
Given: Equation of parabola $y^2=10 x$
Comparing with $y^2=4 a x$, we have $4 a=10$
$
\Rightarrow a=\frac{10}{4}=\frac{5}{2}
$
$\therefore$ Coordinates of focus are $\left(\frac{5}{2}, 0\right)$.
Axis of parabola is $\mathrm{x}$ axis or $y=0$
Equation of the directrix is $x=\frac{-5}{2}$
$
\Rightarrow 2 x+5=0
$
Length of latus rectum $=\frac{4 \times 5}{2}=10$ units
Ex 10.2 Question 6.
$x^2=-9 y$
Answer.
Given: Equation of parabola
$
x^2=-9 y
$
Comparing with $x^2=-4 a y$, we have $4 a=9$
$
\Rightarrow a=\frac{9}{4}
$
$\therefore$ Coordinates of focus are $\left(0, \frac{-9}{4}\right)$.
Axis of parabola is $y$ axis or $x=0$
Equation of the directrix is $y=\frac{9}{4}$
$
\Rightarrow 4 y-9=0
$
Length of latus rectum $=\frac{4 \times 9}{4}=9$ units
In each of the following Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Ex 10.2 Question 7.
Focus $(6,0)$; directrix $x=-6$
Answer.
Given: Directrix : $x=-6$ and $a=6$
Axis of parabola is $x$-axis.
$\therefore$ The required equation of parabola $y^2=4 a x$
$
\begin{aligned}
& \Rightarrow y^2=4 \times 6 x \\
& \Rightarrow y^2=24 x
\end{aligned}
$
Ex 10.2 Question 8.
Focus $(0,-3)$; directrix $y=3$
Answer.
Given: Directrix: $y=3$ and $a=3$
Axis of parabola is $y$ - axis.
$\therefore$ The required equation of parabola $x^2=-4 a y$
$
\begin{aligned}
& \Rightarrow x^2=-4 \times 3 y \\
& \Rightarrow x^2=-12 y
\end{aligned}
$
Ex 10.2 Question 9.
Vertex $(0,0)$; Focus $(3,0)$
Answer.
Given: vertex is at $(0,0), y=0$ and $a=3$
Axis of parabola is $x$-axis.
$\therefore$ The required equation of parabola $y^2=4 a x$
$
\begin{aligned}
& \Rightarrow y^2=4 \times 3 x \\
& \Rightarrow y^2=12 x
\end{aligned}
$
Ex 10.2 Question 10.
Vertex $(0,0)$; Focus $(-2,0)$
Answer.
Given: vertex is at $(0,0), y=0$ and focus is $\mathrm{a}=2$
Axis of parabola is $x$-axis.
$\therefore$ The required equation of parabola
$
\begin{aligned}
& \Rightarrow y^2=-4 a x \\
& \Rightarrow y^2=-8 x
\end{aligned}
$
Ex 10.2 Question 11.
Vertex $(0,0)$ passing through $(2,3)$ and axis is along $x$-axis.
Answer.
Given: vertex $(0,0)$ and axis is along $x$-axis.
$\therefore$ Parabola is of the form $y^2=4 a x$
Since the parabola passes through the point $(2,3)$.
$
\begin{aligned}
& \therefore(3)^2=4 a \times 2 \\
& \Rightarrow 9=8 a \\
& \Rightarrow a=\frac{9}{8}
\end{aligned}
$
$\therefore$ The required equation of parabola $y^2=\frac{4 \times 9}{8} x$
$
\begin{aligned}
& \Rightarrow y^2=\frac{9}{2} x \\
& \Rightarrow 2 y^2=9 x
\end{aligned}
$
Ex 10.2 Question 12.
Vertex $(0,0)$ passing through $(5,2)$ and symmetric with respect to $y$-axis.
Answer.
Given: vertex $(0,0)$ and axis is along $y$ - axis.
$\therefore$ Parabola is of the form $x^2=4 a y$
Since the parabola passes through the point $(5,2)$.
$
\begin{aligned}
& \therefore(5)^2=4 a \times 2 \\
& \Rightarrow 25=8 a \\
& \Rightarrow a=\frac{25}{8}
\end{aligned}
$
$\therefore$ The required equation of parabola $y^2=\frac{4 \times 25}{8} x$
$
\begin{aligned}
& \Rightarrow y^2=\frac{25}{2} x \\
& \Rightarrow 2 y^2=25 x
\end{aligned}
$
