Question

Let the point \( L \) lying in the first quadrant be one end of a latus rectum of the ellipse \[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \] Let \( P \) and \( Q \) be the points where the normal drawn at \( L \) meets the major and minor axes. Then the distance between \( P \) and \( Q \) is:

• A. \( \frac{\sqrt{5}}{4} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{2\sqrt{2}} \)
• D. \( \frac{\sqrt{5}}{2} \)

Hint:

Use geometric properties of ellipse latus rectum and normal equations to find intersections with axes.

Answer 1

The Correct Option is A

Latus rectum endpoints are at \( x = ae \). For the ellipse: \[ a^2 = 4, b^2 = 3 \Rightarrow e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{2} \Rightarrow x = ae = 1 \Rightarrow L = (1, b^2/a = \frac{3}{2}) \] Normal to ellipse at point \( L \) intersects axes. Find equation of normal and substitute \( y = 0 \) and \( x = 0 \) to get \( P \) and \( Q \). Then use distance formula.