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Question: The straight line x + y = \(\sqrt{2}p\)will touch the hyperbola 4x<sup>2</sup> –9y<sup>2</sup> = 36,...

The straight line x + y = 2p\sqrt{2}pwill touch the hyperbola 4x2 –9y2 = 36, if

A

p2 = 2

B

p2 = 5

C

5p2 = 2

D

2p2 = 5

Answer

2p2 = 5

Explanation

Solution

4x2 –9y2 = 36

Ž x29\frac{x^{2}}{9}y24\frac{y^{2}}{4}= 1

line x + y = 2\sqrt{2}p

Ž y = –x + 2\sqrt{2}p

Condition of tangency c = ± a2m2b2\sqrt{a^{2}m^{2} - b^{2}}

Ž2\sqrt{2}p = ± 9(1)24\sqrt{9( - 1)^{2} - 4} Ž 2p2 = 5