Question

P is a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, N is the foot of the perpendicular from P on the transverse axis.The tangent 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.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $P(x_1,y_1)$ be a point on the hyperbola. Then, the coordinates of N are $(x_1,0)$. The equation of the tangent at $(x_1,y_1)$ is
$\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$
This meets X-axis at $T(\frac{a^2}{x_1},0)$
$\therefore$ $OT.ON=\frac{a^2}{x_1}\times x_1=a^2$