Let for two distinct values of p the lines y = x + p touch the ellipse E: x242 + y232 = 1 at the points A and B. Let the line y = x intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to :
- 48
- 20
- 24
- 36
Answer
(3) 24
Solution
y = mx + c touches x2a2 + y2b2 = 1 if c2 = a2m2 + b2
p2 = 16 + 9 = 25 p = 5
Point of contact = (- a2mc, b2c )
A (- 165, 95 ) (p = 5), B ( 165, - 95 ) (p = -5)
y = x in E: x2 ( 116 + 19 ) = 1 x2 = 14425 x = 125
C ( 125, 125 ), D (- 125, - 125 )
AB and CD both pass through the centre O, so they are the diagonals of the quadrilateral.
Area = 12 | AB CD |, with AB = ( 325, - 185 ), CD = (- 245, - 245 )
= 12 | 325 (- 245 ) - (- 185 ) (- 245 ) |
= 12 768 + 43225 = 12 48 = 24
Answer: (3)