<p>(A) The minimum area of triangle formed by the tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) and coordinate axes is:</p>
<p>(a) \(ab\) sq. units</p>
<p>(b) \(\frac{a^2 + b^2}{2}\) sq. units</p>
<p>(c) \(\frac{(a+b)^2}{2}\) sq. units</p>
<p>(d) \(\frac{a^2 + ab + b^2}{3}\) sq. units</p>
Step-by-Step Solution
Key Concept: Find the equation of a tangent to the ellipse at an arbitrary point, determine where it intersects the coordinate axes, calculate the area of the resulting triangle, and minimize using calculus.
<p><strong>Step 1: Find the tangent equation</strong></p><p>For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent at point $(a\cos\theta, b\sin\theta)$ is:</p><p>$$\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1$$</p><p><strong>Step 2: Find intercepts on coordinate axes</strong></p><p>X-intercept (set $y=0$): $x = \frac{a}{\cos\theta}$</p><p>Y-intercept (set $x=0$): $y = \frac{b}{\sin\theta}$</p><p>For intercepts in positive quadrant, we need $0 < \theta < \frac{\pi}{2}$, giving both intercepts positive.</p><p><strong>Step 3: Calculate triangle area</strong></p><p>The triangle formed by the tangent and coordinate axes has vertices at origin $(0,0)$, $(\frac{a}{\cos\theta}, 0)$, and $(0, \frac{b}{\sin\theta})$.</p><p>$$\text{Area} = \frac{1}{2} \cdot \frac{a}{\cos\theta} \cdot \frac{b}{\sin\theta} = \frac{ab}{2\sin\theta\cos\theta} = \frac{ab}{\sin(2\theta)}$$</p><p><strong>Step 4: Minimize the area</strong></p><p>Area $= \frac{ab}{\sin(2\theta)}$ is minimized when $\sin(2\theta)$ is maximum.</p><p>Since $0 < \theta < \frac{\pi}{2}$, we have $0 < 2\theta < \pi$, so $\max(\sin(2\theta)) = 1$ occurs at $2\theta = \frac{\pi}{2}$, i.e., $\theta = \frac{\pi}{4}$.</p><p><strong>Step 5: Calculate minimum area</strong></p><p>$$\text{Minimum Area} = \frac{ab}{1} = ab \text{ sq. units}$$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A