Let P(6,3) be a point on the hyperbola parabola `x^2/a^2-y^2/b^2=1`If the normal at the point intersects the x-axis at (9,0), then the eccentricity of the hyperbola is

Options

A. `sqrt(5//2)`
B. `sqrt(3//2)`
C. `sqrt2`
D. `sqrt3`

Answer

B

Explanation

`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
Differentiating w.r.t. x.
`(dy)/(dx)=(db^(2))/(ya^(2))`
Therefore, the slope of normal at (6, 3) is - `a^(2)//2b^(2)`
The equation of normal is
`(y-3)=(-a^(2))/(2b^(2))(x-6)`
It passes through the point (9, 0). Therefore,
`(a^(2))/(2b^(2))=1or(b^(2))/(a^(2))=(1)/(2)." "thereforee^(2)=a+(b^(2))/(a^(2))=1+(1)/(2)`
`therefore" "e=sqrt((3)/(2))`