<p>For the ellipse \(\dfrac{x^2}{9} + \dfrac{y^2}{5} = 1\), one end of the latus rectum is at \(P(ae, b^2/a)\). Find the equation of the tangent at this point.</p>
Step-by-Step Solution
Key Concept: Use the latus rectum endpoint coordinates with the ellipse tangent formula T = 1, where the point lies on the ellipse and satisfies the focal chord property with a = 3, b² = 5, e = 2/3.
<p><strong>Step 1:</strong> Identify ellipse parameters.</p><p>Given: $\frac{x^2}{9} + \frac{y^2}{5} = 1$</p><p>Here $a^2 = 9$, $b^2 = 5$, so $a = 3$, $b = \sqrt{5}$</p><p>$c^2 = a^2 - b^2 = 9 - 5 = 4$, so $c = 2$</p><p>$e = \frac{c}{a} = \frac{2}{3}$</p><p><strong>Step 2:</strong> Find coordinates of point P.</p><p>$P = (ae, \frac{b^2}{a}) = (3 \cdot \frac{2}{3}, \frac{5}{3}) = (2, \frac{5}{3})$</p><p><strong>Step 3:</strong> Verify P lies on the ellipse.</p><p>$\frac{4}{9} + \frac{25/9}{5} = \frac{4}{9} + \frac{5}{9} = 1$ ✓</p><p><strong>Step 4:</strong> Apply tangent formula.</p><p>For ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, tangent at $(x_1, y_1)$ is: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$</p><p>$\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1$</p><p>$\frac{2x}{9} + \frac{y}{3} = 1$</p><p><strong>Step 5:</strong> Simplify the equation.</p><p>$\frac{2x}{9} + \frac{3y}{9} = 1$</p><p>$2x + 3y = 9$</p><p>∴ Answer: A</p>
Correct Answer: A