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If normal to hyperbola x^2/a^2-y^2/b^2=1...

If normal to hyperbola `x^2/a^2-y^2/b^2=1` drawn at an extremity of its latus-rectum has slope equal to the slope of line which meets hyperbola only once, then the eccentricity of hyperbola is

A

`e = sqrt((1+sqrt(5))/(2))`

B

`e = sqrt((sqrt(5)+3)/(2))`

C

`e = sqrt((2)/(sqrt(5)-1))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Given hyperbola is `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =1`
Differentiate w.r.t.x, we get` (2x)/(a^(2))-(2yy')/(b^(2)) =0` or `y' =(b^(2)x)/(a^(2)y)`
Slope of normal at `(ae,(b^(2))/(a))` is `y' = - (a^(2)(b^(2))/(a))/(b^(2)ae) =- (1)/(e)`
Now line which meets hyperbola only once has slope `+- (b)/(a)` (parallel to asymptote)
According to question `-(b)/(a) =- (1)/(e)`
`rArr (b^(2))/(a^(2)) = (1)/(e^(2))`
`rArr e^(2) -1 =(1)/(e^(2))`
`rArr e^(4) -e^(2) -1 =0`
`rArr e^(2) = (1+- sqrt(5))/(2)`
`rArr e = sqrt((1+sqrt(5))/(2)`
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