Question

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^2}{b^2} + \frac{y^2}{4a^2} = 1$ and the co-ordinate axis is kab, then k is equal to _________.

Hint:

For an ellipse with semi-axes $A$ and $B$, the minimum area of the triangle formed by a tangent and the coordinate axes is always $AB$. Here $A=b$ and $B=2a$, so area is $2ab$.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The equation of a tangent to the ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ in parametric form ($x = A \cos \theta, y = B \sin \theta$) is $\frac{x \cos \theta}{A} + \frac{y \sin \theta}{B} = 1$. The area of the triangle formed with the coordinate axes is $1/2 \times \text{intercept on x-axis} \times \text{intercept on y-axis}$.
Step 2: Detailed Explanation:
For the ellipse $\frac{x^2}{b^2} + \frac{y^2}{(2a)^2} = 1$, the tangent at point $\theta$ is:
\[ \frac{x \cos \theta}{b} + \frac{y \sin \theta}{2a} = 1 \]
Intercept on x-axis (set $y=0$): $x = b/\cos \theta$.
Intercept on y-axis (set $x=0$): $y = 2a/\sin \theta$.
Area of triangle $T = \frac{1}{2} \left| \frac{b}{\cos \theta} \cdot \frac{2a}{\sin \theta} \right| = \frac{ab}{|\sin \theta \cos \theta|}$.
Using $\sin 2\theta = 2 \sin \theta \cos \theta$, we get:
\[ T = \frac{2ab}{|\sin 2\theta|} \]
Area is minimum when $|\sin 2\theta|$ is maximum, i.e., $|\sin 2\theta| = 1$.
Minimum area $= 2ab$.
Comparing with $kab$, we get $k = 2$.
Step 3: Final Answer:
The value of k is 2.