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$\mathrm{P}$ is a point on the hyperbola $\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, N}$ is the foot of the perpendicular from $\mathrm{P}$ on the transverse axis. The tangent to the hyperbola at $\mathrm{P}$ meets the transverse axis at $\mathrm{T}$ . If $\mathrm{O}$ is the centre of the hyperbola, then $\mathrm{OT,ON}$ is equal to
Option: 1
$\mathrm{e^{2}}$

Option: 2
$\mathrm{a^{2}}$

Option: 3
$\mathrm{b^{2}}$

Option: 4
$\mathrm{\frac{b^{2}}{a^{2}}}$

Answers (1)

best_answer

Let $\mathrm{P(x_{1},y_{1})}$ be a point on the hyperbola. Then the co-ordinates of $\mathrm{N}$ are $\mathrm{(x_{1},0)}$
The equation of the tangent at $\mathrm{\left(x_1, y_1\right) \text { is } \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1}$
This meets $\mathrm{x}$ - axis at $\mathrm{T\left(\frac{a^2}{x_1}, 0\right) ; \quad \therefore}$ $\mathrm{\text { OT.ON }=\frac{a^2}{x_1} \times x_1=a^2}$

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