If 6x - 5y - 20 = 0 is a normal to the ellips | vedclass

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(D) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \frac{m(a^2 - b^2)}{\sqrt{a^2 + b^2m^2}}$.Given t

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$37$

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(D) The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \frac{m(a^2 - b^2)}{\sqrt{a^2 + b^2m^2}}$.Given the normal equation $6x - 5y - 20 = 0$,we can rewrite it as $y = \frac{6}{5}x - 4$. Thus,$m = \frac{6}{5}$ and the constant term is $-4$.For the ellipse $x^2 + 3y^2 = k$,we have $\frac{x^2}{k} + \frac{y^2}{k/3} = 1$,so $a^2 = k$ and $b^2 = \frac{k}{3}$.Substituting these into the normal equation constant term: $\frac{m(a^2 - b^2)}{\sqrt{a^2 + b^2m^2}} = 4$.$\frac{\frac{6}{5}(k - \frac{k}{3})}{\sqrt{k + \frac{k}{3} \cdot (\frac{6}{5})^2}} = 4$.$\frac{\frac{6}{5} \cdot \frac{2k}{3}}{\sqrt{k + \frac{k}{3} \cdot \frac{36}{25}}} = 4$ $\Rightarrow \frac{\frac{4k}{5}}{\sqrt{k(1 + \frac{12}{25})}} = 4$.$\frac{4k}{5\sqrt{k} \cdot \sqrt{\frac{37}{25}}} = 4 \Rightarrow \frac{4\sqrt{k}}{5 \cdot \frac{\sqrt{37}}{5}} = 4$.$\frac{\sqrt{k}}{\sqrt{37}} = 1$ $\Rightarrow \sqrt{k} = \sqrt{37}$ $\Rightarrow k = 37$.

If $6x - 5y - 20 = 0$ is a normal to the ellipse $x^2 + 3y^2 = k$,then $k =$

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