"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 ____________"

Question

Accepted Answer

"$\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$

$r^{2}=(x-2)^{2}+y^{2}$

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 $r$

$$ \begin{aligned} & 1+\alpha^{2}=\frac{128}{10} \\\\ & \alpha^{2}=\frac{118}{10} \\\\ &=118.00 \end{aligned} $$"

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 ____________

Solution

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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 ____________

Solution

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