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
A
1
B
2
C
3
D
4
Text Solution
Verified by Experts
The correct Answer is:
C
`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`