If frac{k}{alpha^3} is the length of the subn | vedclass

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(D) Given curve is $x^2-a^2=\frac{x^2 y^2}{a^2}$.Differentiating with respect to $x$:$2x = \frac{1}{a^2} [x^2(2y) \frac{dy}{dx} + (2x)y^2]$.Dividing b

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$a^4$

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(D) Given curve is $x^2-a^2=\frac{x^2 y^2}{a^2}$.Differentiating with respect to $x$:$2x = \frac{1}{a^2} [x^2(2y) \frac{dy}{dx} + (2x)y^2]$.Dividing by $2x$ (assuming $x 
eq 0$):$1 = \frac{1}{a^2} [xy \frac{dy}{dx} + y^2]$.$a^2 = xy \frac{dy}{dx} + y^2 \Rightarrow xy \frac{dy}{dx} = a^2 - y^2$.Thus, $\frac{dy}{dx} = \frac{a^2 - y^2}{xy}$.At point $P(\alpha, y)$, the slope is $\frac{dy}{dx} = \frac{a^2 - y^2}{\alpha y}$.The length of the subnormal is $|y \frac{dy}{dx}| = |y \cdot \frac{a^2 - y^2}{\alpha y}| = |\frac{a^2 - y^2}{\alpha}|$.Since $P(\alpha, y)$ lies on the curve, $\alpha^2 - a^2 = \frac{\alpha^2 y^2}{a^2} \Rightarrow y^2 = \frac{a^2}{\alpha^2}(\alpha^2 - a^2) = a^2 - \frac{a^4}{\alpha^2}$.Substituting $y^2$ into the subnormal formula:Length $= \frac{a^2 - (a^2 - \frac{a^4}{\alpha^2})}{\alpha} = \frac{a^4}{\alpha^3}$.Comparing with $\frac{k}{\alpha^3}$, we get $k = a^4$.

If $\frac{k}{\alpha^3}$ is the length of the subnormal at any point $P(\alpha, y)$ on the curve $x^2-a^2=\frac{x^2 y^2}{a^2}$, then $k=$

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