If a tangent of slope 2 of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the point $(-2, 0)$, then the value of $a^2$ is equal to
Step-by-Step Solution
Key Concept: The equation of a tangent to an ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$; using the point through which it passes determines the ellipse parameters.
A tangent of slope 2 to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ has equation $y = 2x \pm \sqrt{4a^2 + b^2}$. Since it passes through $(-2, 0)$, we have $0 = 2(-2) \pm \sqrt{4a^2 + b^2}$, so $\sqrt{4a^2 + b^2} = 4$, giving $4a^2 + b^2 = 16$. From the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and solving: $a^2 = \frac{15}{4}$, so $a^2 = 3.75$.
Correct Answer: 3.75