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Q.
If $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ touches the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then its eccentric angle $\theta$ is equal to
Conic Sections
Solution:
Equation of any tangent to the ellipse
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ is
$\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 ...$(1)
Also, $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ touches the given ellipse.
Comparing coefficients in (1) and (2), we get
$ \frac{\cos \theta / a}{1 / a}=\frac{\sin \theta / b}{1 / b}=\frac{1}{\sqrt{2}} $
$\Rightarrow \cos \theta=\frac{1}{\sqrt{2}}=\sin \theta $
$\therefore \theta=45^{\circ}$