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Q.
If $(\sqrt{3}) bx + ay -2 ab$ touches the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, then the eccentric angle of the point of contact is
Conic Sections
Solution:
The equation of tangent is
$\frac{ x }{ a } \frac{\sqrt{3}}{2}+\frac{ y }{ b } \frac{1}{2}=1....$(i)
and the equation of tangent at the point $(a \cos \phi, b$ $\sin \phi)$ is
$\frac{x}{a} \cos \phi+\frac{y}{b} \sin \phi=1...$(ii)
Comparing (i) and (ii), we have
$\cos \phi=\frac{\sqrt{3}}{2}$ and $\sin \phi=\frac{1}{2}$
Hence, $\phi=\pi / 6$
