Get Answers to all your Questions

header-bg qa

\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}  

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE