JEE MAIN - Mathematics (2023 - 24th January Morning Shift - No. 17)
Let C be the largest circle centred at (2, 0) and inscribed in the ellipse $${{{x^2}} \over {36}} + {{{y^2}} \over {16}} = 1$$. If (1, $$\alpha$$) lies on C, then 10 $$\alpha^2$$ is equal to ____________
Answer
118
Explanation
$\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$
$r^{2}=(x-2)^{2}+y^{2}$
Solving simultaneously
$-5 x^{2}+36 x+\left(9 r^{2}-180\right)=0$
$D=0$
$r^{2}=\frac{128}{10}$
Distance between $(1, \alpha)$ and $(2,0)$ should be $r$
$$ \begin{aligned} & 1+\alpha^{2}=\frac{128}{10} \\\\ & \alpha^{2}=\frac{118}{10} \\\\ &=118.00 \end{aligned} $$
$r^{2}=(x-2)^{2}+y^{2}$
Solving simultaneously
$-5 x^{2}+36 x+\left(9 r^{2}-180\right)=0$
$D=0$
$r^{2}=\frac{128}{10}$
Distance between $(1, \alpha)$ and $(2,0)$ should be $r$
$$ \begin{aligned} & 1+\alpha^{2}=\frac{128}{10} \\\\ & \alpha^{2}=\frac{118}{10} \\\\ &=118.00 \end{aligned} $$
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