Ellipse Ques 15
- If the normal to the ellipse $3 x^{2}+4 y^{2}=12$ at a point $P$ on it is parallel to the line, $2 x+y=4$ and the tangent to the ellipse at $P$ passes through $Q(4,4)$ then $P Q$ is equal to
(a) $\frac{5 \sqrt{5}}{2}$
(b) $\frac{\sqrt{61}}{2}$
(c) $\frac{\sqrt{221}}{2}$
(d) $\frac{\sqrt{157}}{2}$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 15.(a)
Solution:
Formula:
- Key Idea Equation of tangent and normal to the ellipse
$ \begin{aligned} & \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { at point } p\left(x _1, y _1\right) \text { is } T=0 \Rightarrow \frac{x x _1}{a^{2}}+\frac{y y _1}{b^{2}}=1 \\ & \text { and } \frac{a^{2} x}{x _1}-\frac{b^{2} y}{y _1}=a^{2}-b^{2} \text { respectively. } \end{aligned} $
Equation of given ellipse is $3 x^{2}+4 y^{2}=12$
$\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$……(i)
Now, let point $P(2 \cos \theta, \sqrt{3} \sin \theta)$, so equation of tangent to ellipse (i) at point $P$ is
$ \frac{x \cos \theta}{2}+\frac{y \sin \theta}{\sqrt{3}}=1 $……(ii)
Since, tangent (ii) passes through point $Q(4,4)$
$ \therefore 2 \cos \theta+\frac{4}{\sqrt{3}} \sin \theta=1 $……(iii)
and equation of normal to ellipse (i) at point $P$ is
$ \begin{aligned} & \frac{4 x}{2 \cos \theta}-\frac{3 y}{\sqrt{3} \sin \theta}=4-3 \\ \Rightarrow \quad & 2 x \sin \theta-\sqrt{3} \cos \theta y=\sin \theta \cos \theta \end{aligned} $……(iv)
Since, normal (iv) is parallel to line, $2 x+y=4$
$\therefore$ Slope of normal (iv) $=$ slope of line, $2 x+y=4$
$\Rightarrow \quad \frac{2}{\sqrt{3}} \tan \theta=-2 \Rightarrow \tan \theta=-\sqrt{3} \Rightarrow \theta=120^{\circ}$
$\Rightarrow(\sin \theta, \cos \theta)=(\frac{\sqrt{3}}{2},-\frac{1}{2})$
Hence, point $P(-1, \frac{3}{2})$
Now, $P Q=\sqrt{(4+1)^{2}+(4-\frac{3}{2})^{2}}$ [given cordinates of $Q \equiv(4,4)$ ]
$ =\sqrt{25+\frac{25}{4}}=\frac{5 \sqrt{5}}{2} $
BETA
