The tangent at a point `P` on the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` passes through the point `(0,-b)` and the normal at `P` passes through the point `(2asqrt(2),0)` . Then the eccentricity of the hyperbola is

Options

A. a. 2
B. b. `sqrt2`
C. c. 3
D. d. `sqrt3`

Answer

B

Explanation

The equation of tangent at `(x_(1),y_(1))` is
`(x x_(1))/(a^(2))-(yy_(1))/(b^(2))=1`
It passes through `(0, -b)`. So,
`0+(y_(1))/(b)=1" or "y_(1)=b`
The equation of normal is
`(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)e^(2)`
which passes through `(2asqrt2,0)`. Hence,
`(a^(2)2asqrt2)/(x_(1))=a^(2)e^(2)`
`"or "x_(1)=(2asqrt2)/(e^(2))` ltBrgt Now, `(x_(1),y_(1))` lies on the hyperbola. Therefore,
`(x_(1)^(2))/(a^(2))-(y_(1)^(2))/(b^(2))=1`
`"or "(8)/(e^(4))-1=1`
`"or "e^(2)=2`