The area (in sq. units) of the triangle formed by the tangent and the normal at the point…
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
$\frac{a}{b}\left(a^2+b^2\right)$
4ab
$\frac{b}{4 a}\left(a^2+b^2\right)$
2ab
Solution
Given curve,
$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$
Differentiate w.r.t $x$, we get
$\Rightarrow \quad \frac{2 x}{a^2}+\frac{2 y \cdot y^{\prime}}{b^2}=0 \Rightarrow y^{\prime}=\frac{-b^2 x}{y a^2}$
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, $\quad m_1=-\frac{b}{a}$
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) .
$
