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Question

$If a tangent of slope 2 of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 is normal to the circle x^{2} + y^{2} + 4 x + 1 = 0, then the maximum value of a b is$

Moderate

Solution

$Equation of tangent of \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 with slope m is$
$\begin{aligned} y = m x \pm \sqrt{a^{2} m^{2} + b^{2}} \\ \Rightarrow y = 2 x \pm \sqrt{4 a^{2} + b^{2}} \end{aligned}$
$It is normal to the circle x^{2} + y^{2} + 4 x + 1 = 0 hence, this tangent$
$\begin{array}{l} passes through centre (- 2, 0) \\ \Rightarrow 0 = - 4 \pm \sqrt{4 a^{2} + b^{2}} \\ \Rightarrow 4 a^{2} + b^{2} = 16 \end{array}$
$Using Arithmetic mean (A.M.) \geq Geometric mean (G.M.)$
$\begin{array}{l} we get \frac{4 a^{2} + b^{2}}{2} \geq \sqrt{4 a^{2} b^{2}} \\ \Rightarrow a b \leq 4 \end{array}$
$Minimum value of ab is 4 Therefore, the correct answer is 0004.00$

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