If the line frac{x}{a} + frac{y}{b} = sqrt{2} | vedclass

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(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point with eccentric angle $	heta$ is given by $\frac{x

Vedclass · Accepted Answer

$45$

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(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point with eccentric angle $	heta$ is given by $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Comparing this with the given equation $\frac{x}{a} + \frac{y}{b} = \sqrt{2}$,we rewrite the given equation as $\frac{x}{a \sqrt{2}} + \frac{y}{b \sqrt{2}} = 1$.Equating the coefficients,we get $\cos 	heta = \frac{1}{\sqrt{2}}$ and $\sin 	heta = \frac{1}{\sqrt{2}}$.This implies $	heta = \frac{\pi}{4}$,which is $45^{\circ}$.

If the line $\frac{x}{a} + \frac{y}{b} = \sqrt{2}$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then its eccentric angle $\theta = ............^{\circ}$

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