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JEE Main20253 Apr 2025Morning ShiftMathematicsEllipseActual

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

Options

  1. A218
  2. B377
  3. C290
  4. D338

Correct answer

D. 338

Step-by-step solution

Given ellipse is x^2 36 + y^2 25 =1 Any point on line A B can be assumed as Q ( 5 + r , 5 + r ) Putting this in equation of ellipse, we get 25( 5 +r )^2+36( 5 +r )^2=900 Simplifying, we get aligned & r ^2 (25 ^2 +36 ^2 )+2 5 r (25 +36 )-595=0 & | r |= PA , ~PB aligned Thus, aligned PA PB & = 595 25 ^2 +36 ^2 = 595 25+11 ^2 & = maximum, if ^2 =0 aligned This means line A B must be parallel to x -axis y_A=y_B= 5 Putting y = 5 in equation of ellipse, we get x^2 36 + 1 5 =1 x^2=36 4 5 Hence, PA ^2+ PB ^2= ( 5 - 12 5 )^

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