<p>If the tangent at a point on the ellipse \(\dfrac{x^2}{27} + \dfrac{y^2}{3} = 1\) meets the coordinates axes at \(A\) and \(B\), and \(O\) is the origin, then the minimum area (in sq. units) of the triangle \(OAB\) is</p>
Step-by-Step Solution
Key Concept: For a point on the ellipse, the tangent line equation is derived using the standard form, then intercepts on coordinate axes are found. The area of triangle OAB equals (1/2)|OA||OB|, which can be expressed as a function of the parameter and minimized using calculus or AM-GM inequality.
<p><strong>Step 1:</strong> For the ellipse $\frac{x^2}{27} + \frac{y^2}{3} = 1$, the tangent at point $(x_0, y_0)$ is: $\frac{xx_0}{27} + \frac{yy_0}{3} = 1$</p><p><strong>Step 2:</strong> Find intercepts with axes:<br>• At x-axis (y=0): $A = (\frac{27}{x_0}, 0)$, so $|OA| = \frac{27}{|x_0|}$<br>• At y-axis (x=0): $B = (0, \frac{3}{y_0})$, so $|OB| = \frac{3}{|y_0|}$</p><p><strong>Step 3:</strong> Area of triangle OAB: $\Delta = \frac{1}{2}|OA||OB| = \frac{1}{2} \cdot \frac{27}{|x_0|} \cdot \frac{3}{|y_0|} = \frac{81}{2|x_0y_0|}$</p><p><strong>Step 4:</strong> Since $(x_0, y_0)$ lies on the ellipse: $\frac{x_0^2}{27} + \frac{y_0^2}{3} = 1$<br>By AM-GM inequality: $\frac{x_0^2}{27} + \frac{y_0^2}{3} \geq 2\sqrt{\frac{x_0^2 y_0^2}{81}} = \frac{2|x_0y_0|}{9}$<br>Therefore: $1 \geq \frac{2|x_0y_0|}{9}$, which gives $|x_0y_0| \leq \frac{9}{2}$</p><p><strong>Step 5:</strong> Thus minimum area: $\Delta_{min} = \frac{81}{2 \cdot \frac{9}{2}} = \frac{81}{9} = 9$ sq. units<br>Equality holds when $\frac{x_0^2}{27} = \frac{y_0^2}{3}$, giving $|x_0y_0| = \frac{9}{2}$</p><p>∴ Answer: <strong>C (9 sq. units)</strong></p>
Correct Answer: C