Question

The area (in sq. units) of the triangle formed by the tangent and the normal at the point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ to the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the $X$ -axis is

• A. $\frac{a}{b}\left(a^{2}+b^{2}\right)$
• B. $4 a b$
• C. $\frac{b}{4 a}\left(a^{2}+b^{2}\right)$
• D. $2 a b$

Answer 1

Answer: (C)

Given curve,
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Differentiate w.r.t $x$, we get
$\Rightarrow \frac{2 x}{a^{2}}+\frac{2 y \cdot y^{'}}{b^{2}}=0$
$ \Rightarrow y^{'}=\frac{-b^{2} x}{y a^{2}}$
At $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right), y^{'}=\frac{-b^{2}\left(\frac{a}{\sqrt{2}}\right)}{\frac{b}{\sqrt{2}}\left(a^{2}\right)}=-\frac{b}{a}$
So, slope of tangent, $m_{1}=-\frac{b}{a}$
and slope of normal, $m_{2}=\frac{a}{b}$
Equation of tangent at $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is
$\left(y-\frac{b}{\sqrt{2}}\right)=\frac{-b}{a}\left(x-\frac{a}{\sqrt{2}}\right)$
And equation of normal at $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is
$\left(y-\frac{b}{\sqrt{2}}\right)=\frac{a}{b}\left(x-\frac{a}{\sqrt{2}}\right)$
Now,

Area of $\Delta=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2}\left(\frac{a^{2}+b^{2}}{\sqrt{2} a}\right) \times \frac{b}{\sqrt{2}}=\frac{b}{4 a}\left(a^{2}+b^{2}\right)$