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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

A

`e^(2)`

B

`a^(2)`

C

`b^(2)`

D

`b^(2)//a^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B


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)`
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  17. Portion of asymptote of hyperbola x^2/a^2-y^2/b^2 = 1 (between centre ...

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