Question

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 $(1, a)$ lies on $C$, then $10 \alpha^2$ is equal to

Answer 1

Correct Answer: 118

The correct answer is 118.

 

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
$3\sec \theta x-2\mathrm{cosec}\theta y=10$ 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 $P=\left(\frac{18}{5},\frac{16}{5}\right)$
So the largest radius of circle
$r=\frac{\sqrt{320}}{5}$
So the equation of circle $(x-2{)}^2+y^2=\frac{64}{5}$
Passing it through $(1,\alpha )$
Then ${\alpha }^2=\frac{59}{5}$
$10{\alpha }^2=118$

Answer 2

The equation of the normal to the ellipse at any point \( P = (6 \cos \theta, 4 \sin \theta) \) is: \[ 3 \sec \theta \cdot x - 2 \csc \theta \cdot y = 10. \] This normal is also the normal to the circle passing through the point \( (2, 0) \). Substituting \( (2, 0) \) into the equation: \[ 6 \sec \theta = 10 \quad \text{or} \quad \sin \theta = 0 \quad \text{(not possible)}. \] Thus: \[ \cos \theta = \frac{3}{5}, \quad \sin \theta = \frac{4}{5}. \] The point on the ellipse is: \[ P = \left( 18/5, 16/5 \right). \] The largest radius of the circle is: \[ r = \sqrt{\frac{64}{5}}. \] The equation of the circle becomes: \[ (x - 2)^2 + y^2 = \frac{64}{5}. \] Passing through \( (1, \alpha) \), we have: \[ (1 - 2)^2 + \alpha^2 = \frac{64}{5}. \] Simplifying: \[ 1 + \alpha^2 = \frac{64}{5} \implies \alpha^2 = \frac{59}{5}. \] Thus: \[ 10\alpha^2 = 10 \cdot \frac{59}{5} = 118. \] Final Answer: 118.