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

Answer 1

Answer: (C)

Given curve,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Differentiate w.r.t $x$, we get
$\Rightarrow \frac{2x}{a^2}+\frac{2y\cdot y^{{}^{\prime }}}{b^2}=0$
$\Rightarrow y^{{}^{\prime }}=\frac{-b^{2}x}{y{a}^2}$
At $\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right),y^{{}^{\prime }}=\frac{-b^2\left(\frac{a}{\sqrt{2}}\right)}{\frac{b}{\sqrt{2}}\left(a^2\right)}=-\frac{b}{a}$
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
$=\frac{1}{2}\left(\frac{a^2+b^2}{\sqrt{2}a}\right)\times \frac{b}{\sqrt{2}}=\frac{b}{4a}\left(a^2+b^2\right)$