Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The tangent at a point P on the hyperbola<img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac

The tangent at a point P on the hyperbola\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} meets one of the directrix in F. If PF subtends
an angle \mathrm{\theta } at the corresponding focus, then \mathrm{\theta } equals

Option: 1

\mathrm{\pi / 4}


Option: 2

\mathrm{\pi / 2}


Option: 3

\mathrm{3\pi / 4}


Option: 4

\mathrm{\pi }


Answers (1)

best_answer

Let directrix be \mathrm{x=a / e} and focus be \mathrm{S(ae, 0).}Let \mathrm{P(a \sec \theta, b \tan \theta)} be any point on the curve.
Equation of tangent at P is\mathrm{\frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1} Let F be the intersection point of tangent of directrix,
Then \mathrm{\mathrm{F}=\left(\mathrm{a} / \mathrm{e}, \frac{\mathrm{b}(\sec \theta-\mathrm{e})}{\mathrm{e} \tan \theta}\right)}
\mathrm{\Rightarrow \quad \mathrm{m}_{\mathrm{SF}}=\frac{\mathrm{b}(\sec \theta-\mathrm{e})}{-\mathrm{e} \tan \theta\left(\mathrm{a}^2-1\right)}, \mathrm{m}_{\mathrm{PS}}=\frac{\mathrm{b} \tan \theta}{\mathrm{a}(\sec \theta-\mathrm{e})} \Rightarrow \mathrm{m}_{\mathrm{SF}} \cdot \mathrm{m}_{\mathrm{PS}}=-1}
Hence (b) is the correct answer.

Posted by

mansi

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year
anam-khan

BTech Admission 2025 Live: JEE Mains registrations closing tomorrow; VITEEE, MET exam dates out
anam-khan

JEE Main 2025 Live: NTA to close JEE registration on Friday; apply at jeemain.nta.nic.in, correction window

Latest Question