Miscellaneous Exercise (Revised) - Chapter 10 Conic Sections class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 10: Conic Sections Class 11 NCERT Solutions - Maths

Example 1

The focus of a parabolic mirror as shown in Fig 10.31 is at a distance of $5 \mathrm{~cm}$ from its vertex. If the mirror is $45 \mathrm{~cm}$ deep, find the distance $\mathrm{AB}$ 

Solution

Since the distance from the focus to the vertex is $5 \mathrm{~cm}$. We have, $a=5$. If the origin is taken at the vertex and the axis of the mirror lies along the positive $x$-axis, the equation of the parabolic section is

Note that $y^2=4$ (5) $x=20 x$ $x=45$. Thus $y^2=900$

Therefore
$
\begin{aligned}
& y^2=4(5) x=20 x \\
& x=45 . \text { Thus } \\
& y^2=900 \\
& y= \pm 30 \\
& \mathrm{AB}= 2 y=2 \times 30=60 \mathrm{~cm} .
\end{aligned}
$

Hence
$
\begin{aligned}
& y^2=4(5) x=20 x \\
& x=45 . \text { Thus } \\
& y^2=900 \\
& y= \pm 30 \\
& \mathrm{AB}= 2 y=2 \times 30=60 \mathrm{~cm} .
\end{aligned}
$

Example 2

$\mathrm{~A}$ beam is supported at its ends by
Fig 10.31 supports which are 12 metres apart. Since the load is concentrated at its centre, there is a deflection of $3 \mathrm{~cm}$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \mathrm{~cm}$ ?
Solution Let the vertex be at the lowest point and the axis vertical. Let the coordinate axis be chosen as shown in Fig 10.32.

The equation of the parabola takes the form $x^2=4 a y$. Since it passes through $\left(6, \frac{3}{100}\right)$, we have $(6)^2=4 a\left(\frac{3}{100}\right)$, i.e., $a=\frac{36 \times 100}{12}=300 \mathrm{~m}$

Let $\mathrm{AB}$ be the deflection of the beam which is $\frac{1}{100} \mathrm{~m}$. Coordinates of $\mathrm{B}$ are $\left(x, \frac{2}{100}\right)$.
Therefore
$
\begin{aligned}
x^2 & =4 \times 300 \times \frac{2}{100}=24 \\
x & =\sqrt{24}=2 \sqrt{6} \text { metres }
\end{aligned}
$

Example 3

A rod $\mathrm{AB}$ of length $15 \mathrm{~cm}$ rests in between two coordinate axes in such a way that the end point $\mathrm{A}$ lies on $x$-axis and end point $\mathrm{B}$ lies on $y$-axis. A point $\mathrm{P}(x, y)$ is taken on the rod in such a way that $\mathrm{AP}=6 \mathrm{~cm}$. Show that the locus of $\mathrm{P}$ is an ellipse.
Solution

Let $\mathrm{AB}$ be the rod making an angle $\theta$ with $\mathrm{OX}$ as shown in Fig 10.33 and $\mathrm{P}(x, y)$ the point on it such that $\mathrm{AP}=6 \mathrm{~cm}$.
Since
$
\begin{aligned}
\mathrm{AB} & =15 \mathrm{~cm}, \text { we have } \\
\mathrm{PB} & =9 \mathrm{~cm} .
\end{aligned}
$

From $\mathrm{P}$ draw PQ and PR perpendiculars on $y$-axis and $x$-axis, respectively.

From
$
\Delta \mathrm{PBQ}, \cos \theta=\frac{x}{9}
$

From
$
\triangle \text { PRA, } \sin \theta=\frac{y}{6}
$

Since $\cos ^2 \theta+\sin ^2 \theta=1$
or
$
\left(\frac{x}{9}\right)^2+\left(\frac{y}{6}\right)^2=1
$

Thus the locus of $\mathrm{P}$ is an ellipse.