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If \mathrm{\frac{x}{a}-\frac{y}{b}=\sqrt{2}} touches the ellipse \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}, then eccentric angle of the point of contact is equal to

Option: 1

\frac{\pi}{4}


Option: 2

\frac{3 \pi}{4}


Option: 3

\frac{5 \pi}{4}


Option: 4

\frac{7 \pi}{4}


Answers (1)

best_answer

If P \equiv(a \cos \theta, b \sin \theta) is the point of contact, then the equation of the tangent is

\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1.Comparing we get

\mathrm{\cos \theta=\frac{1}{\sqrt{2}}, \sin \theta=-\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{7 \pi}{4}}.

Posted by

Devendra Khairwa

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