Question

If the normal drawn at the point $$ P \left(\frac{\pi}{4}\right) $$ on the ellipse $$ x^2 + 4y^2 - 4 = 0 $$ meets the ellipse again at $ Q(\alpha, \beta) $, then find $ \alpha $.

• A. \( \sqrt{2} \)
• B. \( \frac{-23}{17\sqrt{2}} \)
• C. \( \frac{7\sqrt{2}}{17} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

Use parametric equations and normal formulas for ellipse to find intersection points.

Answer 1

The Correct Option is C

Parametric form of ellipse: \[ x = a \cos \theta, \quad y = b \sin \theta, \quad a=2, b=1 \] Point \( P \) corresponds to \( \theta = \frac{\pi}{4} \). Equation of normal at \( P \): \[ a^2 y = b^2 x \tan \theta - a b^2 \sin \theta \] Find \( Q \) by solving intersection of normal and ellipse. Calculate \( \alpha \) from solution.