Miscellaneous Exercise (Revised) - Chapter 11 - Conic Sections - Ncert Solutions class 11 - Maths

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Chapter 10: Conic Sections Class 11 NCERT Solutions - Maths

Miscellaneous Exercise Question 1.

If a parabolic reflector is $20 \mathrm{~cm}$ in diameter and $5 \mathrm{~cm}$ deep, find the focus.

Answer.

A parabolic reflector with diameter $\mathrm{PR}=20 \mathrm{~cm}$ and $\mathrm{OQ}=5 \mathrm{~cm}$
Vertex of parabola is $(0,0)$
Let focus of the parabola be $(a, 0)$.

Now, $\mathrm{PR}=20 \mathrm{~cm}$
$
\Rightarrow \mathrm{PQ}=10 \mathrm{~cm}
$
$\therefore$ Coordinate of the point $\mathrm{P}$ are $(5,10)$
Since the point $\mathrm{P}$ lies on the parabola $y^2=4 a x$.
$
\begin{aligned}
& \therefore(10)^2=4 a \times 5 \\
& \Rightarrow a=\frac{100}{20}=5
\end{aligned}
$

Hence the focus i.e $(5,0)$ is the mid point of the given diameter.

Miscellaneous Exercise Question 2.

An arch is in the form of a parabola with its axis vertical. The arch is $10 \mathrm{~m}$ high and 5 $\mathrm{m}$ wide at the base. How wide is it $2 \mathrm{~m}$ from the vertex of the parabola?

Answer.

Let $\mathrm{AB}$ be the parabolic arch having $\mathrm{O}$ as the vertex and $\mathrm{OY}$ as the axis.

The parabola is of the form $x^2=4 a y$.

Now, $C D=5 \mathrm{~m}$
$
\Rightarrow \mathrm{OD}=2.5 \mathrm{~m}
$

And $\mathrm{BD}=10 \mathrm{~m}$

Therefore, coordinates of point B are (2.5, 10).
Since the point B lies on the parabola $x^2=4 a y$.
$
\begin{aligned}
& \therefore(2.5)^2=4 a \times 10 \\
& \Rightarrow a=\frac{6.25}{40} \\
& \Rightarrow a=\frac{5}{32}
\end{aligned}
$
$\therefore$ Equation of the parabola is $x^2=4 \times \frac{5}{32} y$
$
\Rightarrow x^2=\frac{5}{8} y
$

Let $\mathrm{PQ}=d$
$
\Rightarrow \mathrm{NQ}=\frac{d}{2}
$
$\therefore$ Coordinate of the point $Q$ are $\left(\frac{d}{2}, 2\right)$
Since the point Q lies on the parabola $x^2=\frac{5}{8} y$
$
\begin{aligned}
& \therefore\left(\frac{d}{2}\right)^2=\frac{5}{8} \times 2 \\
& \Rightarrow \frac{d^2}{4}=\frac{5}{4} \\
& \Rightarrow d^2=5 \\
& \Rightarrow d=\sqrt{5}
\end{aligned}
$

Therefore, width of arch is $\sqrt{5} \mathrm{~m}=2.24 \mathrm{~m}$ approx.

Miscellaneous Exercise Question 3.

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100 \mathrm{~m}$ long is supported by vertical wires attached to the cable, the longest wire being $30 \mathrm{~m}$ and the shortest being $6 \mathrm{~m}$. Find the length of a supporting wire attached to the roadway $18 \mathrm{~m}$ from the middle.

Answer.

Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length $30 \mathrm{~m}$ each. Let $\mathrm{OC}$ be the shortest wire of length $6 \mathrm{~m}$ and $\mathrm{LM}$ be the roadway.

Now $\mathrm{AL}=\mathrm{BM}=30 \mathrm{~m}, \mathrm{OC}=6 \mathrm{~m}$ and $\mathrm{LM}=100 \mathrm{~m}$

(in figure LM is roadway not roundway) 

$
\therefore \mathrm{LC}=\mathrm{CM}=\frac{1}{2} \mathrm{LM}=50 \mathrm{~m}
$

Let $O$ be the vertex and axis of the parabola be $y$ - axis.
Therefore, the equation of the parabola in standard form is $x^2=4 a y$.
$\therefore$ Coordinates of the point B are $(50,24)$
Since point B lies on the parabola $x^2=4 a y$.
$
\begin{aligned}
& \therefore(50)^2=4 a \times 24 \\
& \Rightarrow a=\frac{2500}{4 \times 24}=\frac{625}{24}
\end{aligned}
$

Therefore, equation of parabola is $x^2=\frac{4 \times 625}{24} y$
$
\Rightarrow x^2=\frac{625}{6} y
$

Let length of the supporting wire PQ at a distance of $18 \mathrm{~m}$ be $h$.
$
\therefore \mathrm{OR}=18 \mathrm{~m} \text { and } \mathrm{PR}=\mathrm{PQ}-\mathrm{QR}=\mathrm{PQ}-\mathrm{OC}=h-6
$
$\therefore$ Coordinates of point $\mathrm{P}$ are $(18, h-6)$

$\text { Now, since the point } \mathrm{P} \text { lies on parabola } x^2=\frac{625}{6} y$

$
\begin{aligned}
& \therefore(18)^2=\frac{625}{6}(h-6) \\
& \Rightarrow 324 \times 6=625 h-3750 \\
& \Rightarrow 625 h=1944+3750 \\
& \Rightarrow h=\frac{5694}{625}=9.11 \mathrm{~m} \text { approx. }
\end{aligned}
$
Miscellaneous Exercise Question 4.

An arch is in the form of a semi-ellipse. It is $8 \mathrm{~m}$ wide and $2 \mathrm{~m}$ high at the centre. Find the height of the arch at a point $1.5 \mathrm{~m}$ from one end.

Answer.

Given: Width of elliptical arch (AB)
$
=2 a=8 \Rightarrow a=4 \mathrm{~m}
$

Height of the centre (OB) $=b=2 \mathrm{~m}$
The axis of ellipse is $x$-axis.
Therefore, the equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
$
\begin{aligned}
& \therefore \frac{x^2}{(4)^2}+\frac{y^2}{(2)^2}=1 \\
& \Rightarrow \frac{x^2}{16}+\frac{y^2}{4}=1
\end{aligned}
$

Now, $\mathrm{AP}=1.5 \mathrm{~m}$

$
\therefore \mathrm{OP}=\mathrm{OA}-\mathrm{AP}=4-1.5=2.5 \mathrm{~m}
$

Let $\mathrm{PQ}=h$, then coordinates of $\mathrm{Q}$ are $(2.5, h)$
Since the point $Q$ lies on the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$
$
\begin{aligned}
& \therefore \frac{(2.5)^2}{16}+\frac{h^2}{4}=1 \\
& =>\frac{6.25}{16}+\frac{h^2}{4}=1 \\
& =>\frac{h^2}{4}=1-\frac{6.25}{16} \\
& \Rightarrow h^2=\frac{9.75 \times 4}{16}=\frac{9.75}{4} \\
& \Rightarrow h^2=2.44 \\
& \Rightarrow h=1.56 \mathrm{~m} \text { approx. }
\end{aligned}
$
Miscellaneous Exercise Question 5.

A rod of length $12 \mathrm{~cm}$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3 \mathrm{~cm}$ from the end in contact with the $x$-axis.

Answer.

Let $\mathrm{AB}$ be a rod of length $12 \mathrm{~cm}$ and $\mathrm{P}(x, y)$ be any point on the rod such that $\mathrm{PA}=3$ $\mathrm{cm}$ and $\mathrm{PB}=9 \mathrm{~cm}$.

Let $\mathrm{AR}=a$ and $\mathrm{BQ}=b$
Then $\triangle A R P \sim \triangle P Q B$
$
\begin{aligned}
& \therefore \frac{\mathrm{AR}}{\mathrm{PQ}}=\frac{\mathrm{AP}}{\mathrm{PB}} \\
& \Rightarrow \frac{a}{x}=\frac{3}{9} \\
& \Rightarrow 9 a=3 x \\
& \Rightarrow a=\frac{x}{3}
\end{aligned}
$

And $\frac{\mathrm{BQ}}{\mathrm{BP}}=\frac{\mathrm{PR}}{\mathrm{PA}}$
$
\begin{aligned}
& \Rightarrow \frac{b}{9}=\frac{y}{3} \\
& \Rightarrow 3 b=9 y \\
& \Rightarrow b=3 y
\end{aligned}
$

Now, $\mathrm{OA}=\mathrm{OR}+\mathrm{AR}=x+a=x+\frac{x}{3}=\frac{4 x}{3}$
And $\mathrm{OB}=\mathrm{OQ}+\mathrm{BQ}=y+b=y+3 y=4 y$

In right angled triangle AOB, $A B^2=O A^2+O B^2$
$
\begin{aligned}
& \Rightarrow(12)^2=\left(\frac{4 x}{3}\right)^2+(4 y)^2 \\
& \Rightarrow 144=\frac{16 x^2}{9}+16 y^2
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{16 x^2}{9 \times 144}+\frac{16 y^2}{144}=1 \\
& \Rightarrow \frac{x^2}{81}+\frac{y^2}{9}=1
\end{aligned}
$
which is required locus of point $\mathrm{P}$ and which represents an ellipse.
Miscellaneous Exercise Question 6.

Find the area of the triangle formed by the lines the vertex of the parabola $x^2=12 y$ to the ends of its latus rectum.

Answer.

Given: Equation of parabola $x^2=12 y$ which is in the form of $x^2=4 a y$

$
\begin{aligned}
& \therefore 4 a=12 \\
& \Rightarrow a=3
\end{aligned}
$

Focus of the parabola is $(0,3)$.
Let $\mathrm{AB}$ be the latus rectum of the parabola, then $y=3$
$
\begin{aligned}
& \therefore x^2=4 \times 3 \times 3=36 \\
& \Rightarrow x= \pm 6
\end{aligned}
$

The coordinates of A are $(-6,3)$
and coordinates of B are $(6,3)$.
$\therefore$ Area of $\triangle O A B$

$
\begin{aligned}
& =\frac{1}{2}|0(3-3)+6(3-0)+(-6)(0-3)| \\
& =\frac{1}{2}|0+18+18| \\
& =\frac{1}{2} \times|36|=18 \text { sq. units }
\end{aligned}
$
Miscellaneous Exercise Question 7.

A man running a race course notes that the sum of the distances from the two flag posts from him is always $10 \mathrm{~m}$ and the distance between the flag posts is $8 \mathrm{~m}$. Find the equation of the posts traced by the man.

Answer.

Let $F_1$ and $F_2$ be two points where the flag parts are fixed on the ground. The origin $\mathrm{O}$ is the mid-point of $F_1 F_2$.

$
\therefore O F_1=O F_2=\frac{1}{2} F_1 F_2=\frac{1}{2} \times 8=4 \mathrm{~m}
$
$\therefore$ Coordinates of $\mathrm{F}_1$ are $(-4,0)$ and $\mathrm{F}_2$ are $(4,0)$.
Let $\mathrm{P}(\alpha, \beta)$ be any point on the track.
$
\begin{aligned}
& \therefore P F_1+P F_2=10 \\
& \Rightarrow \sqrt{(\alpha+4)^2+(\beta-0)^2}+\sqrt{(\alpha-4)^2+(\beta-0)^2}=10 \\
& \Rightarrow \sqrt{\alpha^2+16+8 \alpha+\beta^2}=10-\sqrt{\alpha^2+16-8 \alpha+\beta^2}
\end{aligned}
$

Squaring both sides, we have,

$
\begin{aligned}
& \Rightarrow \alpha^2+\beta^2+8 \alpha+16=100+\alpha^2+\beta^2-8 \alpha+16-20 \sqrt{\alpha^2+16-8 \alpha+\beta^2} \\
& \Rightarrow 16 \alpha-100=-20 \sqrt{\alpha^2+16-8 \alpha+\beta^2}
\end{aligned}
$

Squaring both sides again, we have,
$
\begin{aligned}
& \Rightarrow 256 \alpha^2+10000-3200 \alpha=400\left(\alpha^2+\beta^2-8 \alpha+16\right) \\
& \Rightarrow 256 \alpha^2+10000-3200 \alpha=400 \alpha^2+400 \beta^2-3200 \alpha+6400 \\
& \Rightarrow 144 \alpha^2+400 \beta^2=3600
\end{aligned}
$

Divide both sides of equation by 3600
$
\begin{aligned}
& \Rightarrow \frac{144 \alpha^2}{3600}+\frac{400 \beta^2}{3600}=1 \\
& \Rightarrow \frac{\alpha^2}{25}+\frac{\beta^2}{9}=1 \\
& \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1
\end{aligned}
$
which is the required equation of locus of point $P$.
Miscellaneous Exercise Question 8.

An equilateral triangle is inscribed in the parabola $y^2=4 a x$ where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer.

$\text {Given: Equation of the parabola } y^2=4 a x \text {. }$

Let $b$ be the side of an equilateral $\triangle O A B$ whose one vertex is the vertex of parabola and let $\mathrm{OC}=x$

Now, $\mathrm{AB}=b$
$
\therefore \mathrm{AC}=\mathrm{BC}=\frac{1}{2} \times A B=\frac{b}{2}
$

Coordinates of point A are $\left(x \frac{b}{2}\right)$.
Since, point A lies on the parabola $y^2=4 a x$
$
\begin{aligned}
& \Rightarrow\left(\frac{b}{2}\right)^2=4 a x \\
& \Rightarrow x=\frac{b^2}{4 \times 4 a} \\
& \Rightarrow x=\frac{b^2}{16 a}
\end{aligned}
$

In right angled triangle $\triangle O A C$
$
\mathrm{OA}^2=\mathrm{OC}^2+\mathrm{AC}^2
$

$
\begin{aligned}
& \Rightarrow b^2=x^2+\left(\frac{b}{2}\right)^2 \\
& \Rightarrow b^2=\left(\frac{b^2}{16 a}\right)^2+\frac{b^2}{4} \\
& \Rightarrow b^2=\frac{b^4}{256 a^2}+\frac{b^2}{4} \\
& \Rightarrow 1=\frac{b^2}{256 a^2}+\frac{1}{4} \\
& \Rightarrow \frac{b^2}{256 a^2}=1-\frac{1}{4} \\
& \Rightarrow b^2=\frac{3}{4} \times 256 a^2 \\
& \Rightarrow b^2=192 a^2 \\
& \Rightarrow b=8 \sqrt{3} a
\end{aligned}
$

Therefore, the side of triangle is $8 \sqrt{3} a$.