If x cos alpha + y sin alpha = 4 is a tangent | vedclass

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(D) The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$. Given the

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$	an^{-1}(3/\sqrt{7})$

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(D) The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$. Given the line $x \cos \alpha + y \sin \alpha = 4$,we can rewrite it as $y = -x \cot \alpha + 4 \operatorname{cosec} \alpha$. Here,$m = -\cot \alpha$,$c = 4 \operatorname{cosec} \alpha$,$a^2 = 25$,and $b^2 = 9$. Substituting these into the condition $c^2 = a^2m^2 + b^2$: $16 \operatorname{cosec}^2 \alpha = 25 \cot^2 \alpha + 9$. Using $\operatorname{cosec}^2 \alpha = 1 + \cot^2 \alpha$,we get $16(1 + \cot^2 \alpha) = 25 \cot^2 \alpha + 9$. $16 + 16 \cot^2 \alpha = 25 \cot^2 \alpha + 9$. $9 \cot^2 \alpha = 7 \implies \cot^2 \alpha = 7/9$. Thus,$	an^2 \alpha = 9/7$,which gives $	an \alpha = 3/\sqrt{7}$. Therefore,$\alpha = 	an^{-1}(3/\sqrt{7})$.

If $x \cos \alpha + y \sin \alpha = 4$ is a tangent to $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$,then the value of $\alpha$ is

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