The area of the triangle formed by the tangen | vedclass

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Question

(D) The given curve is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The point $P$ is $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.
Taking th

Vedclass · Accepted Answer

$\frac{b(a^{2}+b^{2})}{4a}$

Vedclass · Answer

(D) The given curve is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The point $P$ is $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}ight)$.
Taking the derivative,$\frac{2x}{a^{2}} + \frac{2y}{b^{2}}\frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{b^{2}x}{a^{2}y}$.
At $P$,the slope of the tangent $m_{t} = -\frac{b^{2}(a/\sqrt{2})}{a^{2}(b/\sqrt{2})} = -\frac{b}{a}$.
The equation of the tangent at $P$ is $y - \frac{b}{\sqrt{2}} = -\frac{b}{a}\left(x - \frac{a}{\sqrt{2}}ight)$,which simplifies to $bx + ay = \sqrt{2}ab$.
For the $x$-axis $(y=0)$,$x = \sqrt{2}a$. So,the first vertex is $A(\sqrt{2}a, 0)$.
The slope of the normal $m_{n} = \frac{a}{b}$.
The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}ight)$,which simplifies to $ax - by = \frac{a^{2}-b^{2}}{\sqrt{2}}$.
For the $x$-axis $(y=0)$,$x = \frac{a^{2}-b^{2}}{a\sqrt{2}}$. So,the second vertex is $B\left(\frac{a^{2}-b^{2}}{a\sqrt{2}}, 0ight)$.
The third vertex is $P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}ight)$.
The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left|\sqrt{2}a - \frac{a^{2}-b^{2}}{a\sqrt{2}}ight| 	imes \frac{b}{\sqrt{2}}$.
$= \frac{1}{2} 	imes \left|\frac{2a^{2}-a^{2}+b^{2}}{a\sqrt{2}}ight| 	imes \frac{b}{\sqrt{2}} = \frac{1}{2} 	imes \frac{a^{2}+b^{2}}{a\sqrt{2}} 	imes \frac{b}{\sqrt{2}} = \frac{b(a^{2}+b^{2})}{4a}$.

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

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