EllipseHardBloom L3
Question
The length of sides of square which can be made by four perpendicular tangents to the ellipse $\frac{x^{2}}{7} + \frac{2y^{2}}{11} = 1$ is:
Options
A.1
B.5
C.6
D.none of these
Solution
Sol.(b) $l^{2} + l^{2} = (AC)^{2} = \left( 2\sqrt{a^{2} + b^{2}} \right)^{2}$
$$(x - 1)^{2} + (y - 1)^{2} = 16 + 9$$
$$\begin{matrix} 2l^{2} & \ = 4\left( 7 + \frac{11}{2} \right) \\ l^{2} & \ = 14 + 11 = 25 \\ \Rightarrow \ l & \ = 5 \end{matrix}$$
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