If a tangent of slope 2 of the ellipse a 2 x 2 + b 2 y 2 = 1 is normal to the circle x 2 + y 2 + 4 x + 1 = 0 then the maximum value of ab is

Question

If a tangent of slope 2 of the ellipse a 2 x 2 ​ + b 2 y 2 ​ = 1 is normal to the circle x 2 + y 2 + 4 x + 1 = 0 then the maximum value of ab is

TestHub · Accepted Answer

The equation of a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with slope $m = 2$ is $y = 2 x \pm 4 a^{2} + b^{2}$ .

The given circle is $x^{2} + y^{2} + 4 x + 1 = 0$ , which can be rewritten as $(x + 2)^{2} + y^{2} = 3$ . Its center is $(- 2, 0)$ .

Since the tangent is normal to the circle, it must pass through the center of the circle.

Substitute $(- 2, 0)$ into the tangent equation: $0 = 2 (- 2) \pm 4 a^{2} + b^{2}$ .

This gives $4 = \pm 4 a^{2} + b^{2}$ , so $16 = 4 a^{2} + b^{2}$ .

By AM-GM inequality, $\frac{4 a ^{2} + b ^{2}}{2} \geq 4 a^{2} b^{2}$ .

$8 \geq 2 ab$ , so $ab \leq 4$ .

The maximum value of $ab$ is 4.

The final answer is $4$ .

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