A tangent having slope $-\frac{1}{8}$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ intersects the major and minor axes at $A$ and $B$. If $O$ is the origin, then the area of $\triangle OAB$ is
Step-by-Step Solution
Key Concept: The area of triangle formed by a tangent to an ellipse and the coordinate axes uses the intercept form and parametric representation of points on the ellipse.
Any point on the ellipse $\frac{x^2}{(a\sqrt{2})^2} + \frac{y^2}{(a\sqrt{2})^2} = 1$ can be parametrized as $(3\sqrt{2}\cos\theta, 4\sqrt{2}\sin\theta)$. The slope of the tangent at this point is $\frac{-b^2x}{a^2y} = \cot\theta$. From the given slope condition $-\frac{1}{4}$ and solving the parametric equations, we determine the specific tangent line. The area enclosed by this tangent and the coordinate axes equals 24 square units.
Correct Answer: 24