P is a point on the hyperbola `(x^(2))/(y^(2))-(y^(2))/(b^(2))=1`, and N is the foot of the perpendicular from P on the transverse axis. The tantent to the hyperbola at P meets the transverse axis at T. If O is the centre of the hyperbola, then OT.ON is equal to

Options

A. `e^(2)`
B. `a^(2)`
C. `b^(2)`
D. `b^(2)//a^(2)`

Answer

B

Explanation

Tangent at point P is
`(x)/(a) sec theta-(y)/(b) tan theta=1`
It meets the x-axis at T `(a cos theta, 0)` and the foot of perpendicular from P to the x-axis is N `(a sec theta, 0)`.
From the diagram,
OT = `a cos theta and ON = a sec theta`
`therefore" "OT*ON=a^(2)`