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Slope Form of Tangent: Ellipse

If a tangent ...

Question

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

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Solution

Equation of tangent is $y = 2 x \pm \sqrt{4 a^{2} + b^{2}}$

If this tangent is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0,$ then it will pass through the centre $(- 2, 0)$ of the circle.
$\Rightarrow 0 = - 4 \pm \sqrt{4 a^{2} + b^{2}} \Rightarrow 4 a^{2} + b^{2} = 16$

Using $A . M . \geq G . M .$ inequality, we get
$\frac{4 a^{2} + b^{2}}{2} \geq \sqrt{4 a^{2} \times b^{2}} \Rightarrow a b \leq 4$

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Slope Form of Tangent: Ellipse

Standard XII Mathematics

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