Ellipse Question 2
Question 2 - 24 January - Shift 1
Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $=\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If $(1, \alpha)$ lies on $C$, then $10 \alpha^{2}$ is equal to
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Answer: 118
Solution:
Formula: Equation of Normal ( Parametric form )
Equation of normal of ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ at any point $P(6 \cos \theta, 4 \sin \theta)$ is
$3 x\sec \theta -2y cosec \theta = 10$ this normal is also the normal of the circle passing through the point $(2,0)$
So,$6 \sec \theta=10$ or $\sin \theta=0$ (Not possible)
$\cos \theta=\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$ so point $P=(\frac{18}{5}, \frac{16}{5})$
So the largest radius of circle is $r=\frac{\sqrt{320}}{5}$
So the equation of circle $(x-2)^{2}+y^{2}=\frac{64}{5}$
Passing it through $(1, \alpha)$
Then $\alpha^{2}=\frac{59}{5}$
$\therefore 10 \alpha^{2}=118$
