Question

A line passing through the point P$(\sqrt{5}, \sqrt{5})$ intersects the ellipse $ \frac{x^2}{36} + \frac{y^2}{25} = 1 $ at A and B such that (PA).(PB) is maximum. Then 5(PA$^2$ + PB$^2$) is equal to :

• A. 218
• B. 377
• C. 290
• D. 338

Hint:

To maximize the product of two distances from a point to the intersection points of a line and an ellipse, the line should be parallel to the major or minor axis of the ellipse.

Answer 1

The Correct Option is D

Given ellipse is \[ \frac{x^2}{36} + \frac{y^2}{25} = 1 \] Any point on line \(AB\) can be assumed as \[ Q(\sqrt{5} + r\cos\theta, \sqrt{5} + r\sin\theta) \] Substituting this into the equation of the ellipse: \[ 25(\sqrt{5} + r\cos\theta)^2 + 36(\sqrt{5} + r\sin\theta)^2 = 900 \] Expanding and simplifying: \[ r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) + 25 \cdot 5 + 36 \cdot 5 = 900 \] \[ r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) = 900 - 305 = 595 \] \[ \Rightarrow r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0 \] Let the roots of this quadratic in \(r\) be \(PA\) and \(PB\), then \[ PA \cdot PB = \frac{595}{25\cos^2\theta + 36\sin^2\theta} \] To maximize \(PA \cdot PB\), the denominator should be minimized: \[ 25\cos^2\theta + 36\sin^2\theta = 25 + 11\sin^2\theta \] Maximum value of \(PA \cdot PB\) occurs when \(\sin^2\theta = 0\), i.e., \(\theta = 0\) or \(\pi\) \[ \Rightarrow \text{Line } AB \text{ must be parallel to the x-axis} \Rightarrow y_A = y_B = \sqrt{5} \] Putting \(y = \sqrt{5}\) in the equation of the ellipse: \[ \frac{x^2}{36} + \frac{5}{25} = 1 \Rightarrow \frac{x^2}{36} = \frac{4}{5} \Rightarrow x^2 = \frac{4}{5} \cdot 36 = \frac{144}{5} \] Hence, coordinates of \(A\) and \(B\) are: \[ A = \left(-\frac{12}{\sqrt{5}}, \sqrt{5}\right), \quad B = \left(\frac{12}{\sqrt{5}}, \sqrt{5} \right) \] Now, \[ PA^2 + PB^2 = \left(\sqrt{5} - \frac{12}{\sqrt{5}}\right)^2 + \left(\sqrt{5} + \frac{12}{\sqrt{5}}\right)^2 \] \[ = 2\left(5 + \frac{144}{5} \right) = 2 \cdot \frac{169}{5} = \frac{338}{5} \] \[ \Rightarrow 5(PA^2 + PB^2) = 338 \]