The length of the sub-tangent to the hyperbola `x^(2)-4y^(2)=4` corresponding to the normal having slope unity is `(1)/(sqrtk),` then the value of k is

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

`x^(2)-4y^(2)=4`
`therefore" "2x-8y y'=0`
Slope of normal is 1.
`therefore" Slope of tangent is "-1.`
`therefore" "x+4y=0`
Also solving (i) and (ii), `16y^(2)-4y^(2)=4`
`therefore" "y^(2)=1//3`
`therefore" Point on the curve is "(pm(4)/(sqrt3),pm(1)/(sqrt3))`
Length of normal `=|y(dx)/(dy)|`
Given `(1)/(sqrtk)=(1)/(sqrt3)`
`therefore" "k=3`