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 :
338
Ellipse Line Intersection Maximization Problem
The problem asks for the value of $ 5(PA^2 + PB^2) $. This is determined when the product of distances $ (PA) \cdot (PB) $ is maximum. The points A and B are where a line through $ P(\sqrt{5}, \sqrt{5}) $ intersects the ellipse $ \frac{x^2}{36} + \frac{y^2}{25} = 1 $.
Solution Strategy
We use the parametric form of the line passing through P. Substituting this into the ellipse equation gives a quadratic equation in the distance parameter $ r $. The product $ (PA) \cdot (PB) $ is related to the roots of this quadratic equation. We find the condition that maximizes this product, identify the corresponding line, and then calculate $ PA^2 + PB^2 $.
Parametric Representation and Quadratic Equation
Let the line through $ P(\sqrt{5}, \sqrt{5}) $ be defined parametrically: $ x = \sqrt{5} + r \cos \theta $ $ y = \sqrt{5} + r \sin \theta $ Substituting into the ellipse equation $ \frac{x^2}{36} + \frac{y^2}{25} = 1 $ and simplifying leads to a quadratic equation in $ r $: $ r^2 (25 \cos^2 \theta + 36 \sin^2 \theta) + r (50\sqrt{5} \cos \theta + 72\sqrt{5} \sin \theta) - 595 = 0 $ Let the roots be $ r_1 $ and $ r_2 $. The distances are $ PA = |r_1| $ and $ PB = |r_2| $.
Maximizing the Distance Product $ (PA) \cdot (PB) $
The product of the roots is $ r_1 r_2 = \frac{-595}{25 \cos^2 \theta + 36 \sin^2 \theta} $. Since P is inside the ellipse, $ r_1 $ and $ r_2 $ have opposite signs. Thus, $ (PA) \cdot (PB) = |r_1 r_2| = \frac{595}{25 \cos^2 \theta + 36 \sin^2 \theta} $. To maximize $ (PA) \cdot (PB) $, we minimize the denominator $ D = 25 \cos^2 \theta + 36 \sin^2 \theta $. Writing $ D = 25 (1 - \sin^2 \theta) + 36 \sin^2 \theta = 25 + 11 \sin^2 \theta $. The minimum value of $ D $ is $ 25 $, achieved when $ \sin^2 \theta = 0 $ (i.e., $ \theta = 0 $ or $ \theta = \pi $). This corresponds to a horizontal line passing through P, with equation $ y = \sqrt{5} $.
Calculating $ PA^2 + PB^2 $
Substitute $ y = \sqrt{5} $ into the ellipse equation: $ \frac{x^2}{36} + \frac{(\sqrt{5})^2}{25} = 1 \implies \frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = 1 - \frac{1}{5} = \frac{4}{5} $ $ x^2 = \frac{36 \times 4}{5} = \frac{144}{5} $. So, $ x = \pm \frac{12}{\sqrt{5}} $. The intersection points are $ A = (\frac{12}{\sqrt{5}}, \sqrt{5}) $ and $ B = (-\frac{12}{\sqrt{5}}, \sqrt{5}) $. The point P is $ (\sqrt{5}, \sqrt{5}) $. $ PA^2 = \left(\frac{12}{\sqrt{5}} - \sqrt{5}\right)^2 + (\sqrt{5} - \sqrt{5})^2 = \left(\frac{12 - 5}{\sqrt{5}}\right)^2 = \left(\frac{7}{\sqrt{5}}\right)^2 = \frac{49}{5} $. $ PB^2 = \left(-\frac{12}{\sqrt{5}} - \sqrt{5}\right)^2 + (\sqrt{5} - \sqrt{5})^2 = \left(\frac{-12 - 5}{\sqrt{5}}\right)^2 = \left(\frac{-17}{\sqrt{5}}\right)^2 = \frac{289}{5} $. $ PA^2 + PB^2 = \frac{49}{5} + \frac{289}{5} = \frac{338}{5} $.
Final Calculation
The required value is $ 5(PA^2 + PB^2) $. $ 5(PA^2 + PB^2) = 5 \times \frac{338}{5} = 338 $.
Similar Questions
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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