The area (in sq. units) of the triangle forme | vedclass

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(C) Given curve: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y y'}{b^2} = 0$,which implie

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

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(C) Given curve: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y y'}{b^2} = 0$,which implies $y' = -\frac{b^2x}{a^2y}$. At the point $P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}ight)$,the slope of the tangent is $m_1 = -\frac{b^2(a/\sqrt{2})}{a^2(b/\sqrt{2})} = -\frac{b}{a}$. The slope of the normal is $m_2 = -\frac{1}{m_1} = \frac{a}{b}$. The equation of the tangent at $P$ is $y - \frac{b}{\sqrt{2}} = -\frac{b}{a}\left(x - \frac{a}{\sqrt{2}}ight)$. Setting $y=0$,we find the $X$-intercept: $-\frac{b}{\sqrt{2}} = -\frac{b}{a}\left(x - \frac{a}{\sqrt{2}}ight)$ $\Rightarrow x = \frac{a}{\sqrt{2}} + \frac{a}{\sqrt{2}} = \frac{2a}{\sqrt{2}} = a\sqrt{2}$. The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}ight)$. Setting $y=0$,we find the $X$-intercept: $-\frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}ight)$ $\Rightarrow x = \frac{a}{\sqrt{2}} - \frac{b^2}{a\sqrt{2}} = \frac{a^2-b^2}{a\sqrt{2}}$. The base of the triangle on the $X$-axis is $|a\sqrt{2} - \frac{a^2-b^2}{a\sqrt{2}}| = |\frac{2a^2 - a^2 + b^2}{a\sqrt{2}}| = \frac{a^2+b^2}{a\sqrt{2}}$. The height of the triangle is the $y$-coordinate of point $P$,which is $\frac{b}{\sqrt{2}}$. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{a^2+b^2}{a\sqrt{2}} 	imes \frac{b}{\sqrt{2}} = \frac{b(a^2+b^2)}{4a}$.

The area (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

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