A line parallel to the straight line 2x-y=0 is tangent to the hyperbola x^2/4 - y^2/2=1 at the point (x1,y1). then (x1)^2+5(y1)^2 is equal to:

Answers (1)

Solution:

\ T :x	imes x1/4-y	imes y1/2=1\ \Rightarrow t:2x-y=0

t is parallel to T

\ T:2x-y=k

Now , \ x1=8/k ; and; y1=2/k

(x1,y1) lies on the  hyperbola

\Rightarrow 64/4k^2-4/2k^2=1\ \Rightarrow k^2=14\ \Rightarrow (x1)^2+5(y1)^2=84/14=6

 

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