Ellipse Ques 26
- The tangent and normal to the ellipse $3 x^{2}+5 y^{2}=32$ at the point $P(2,2)$ meets the $X$-axis at $Q$ and $R$, respectively. Then, the area (in sq units) of the $\triangle P Q R$ is
(a) $\frac{16}{3}$
(b) $\frac{14}{3}$
(c) $\frac{34}{15}$
(d) $\frac{68}{15}$
(2019 Main, 10 April II)
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Answer:
Correct Answer: 26.(d)
Solution:
Formula:
- Equation of given ellipse is
$ 3 x^{2}+5 y^{2}=32 $
Now, the slope of tangent and normal at point $P(2,2)$ to the ellipse (i) are respectively
$ m _T=\left.\frac{d y}{d x}\right| _{(2,2)} \text { and } m _N=-\left.\frac{d x}{d y}\right| _{(2,2)} $
On differentiating ellipse (i), w.r.t. $x$, we get
$ 6 x+10 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{3 x}{5 y} $
So, $m _T=-\left.\frac{3 x}{5 y}\right| _{(2,2)}=-\frac{3}{5}$ and $m _N=\left.\frac{5 y}{3 y}\right| _{(2,2)}=\frac{5}{3}$
Now, equation of tangent and normal to the given ellipse (i) at point $P(2,2)$ are
$ (y-2)=-\frac{3}{5}(x-2) $
and $(y-2)=\frac{5}{3}(x-2)$ respectively.
It is given that point of intersection of tangent and normal are $Q$ and $R$ at $X$-axis respectively.
So, $\quad Q (\frac{16}{3}, 0)$ and $R (\frac{4}{5}, 0)$
$\therefore$ Area of $\triangle P Q R=\frac{1}{2}(Q R) \times$ height
$ \begin{gathered} =\frac{1}{2} \times \frac{68}{15} \times 2=\frac{68}{15} \text { sq units } \\ {\left[\because Q R=\sqrt{(\frac{16}{3}-\frac{4}{5})^{2}}=\sqrt{(\frac{68}{15}})^2=\frac{68}{15} \text { and height }=2\right]} \end{gathered} $
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