Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let a tangent to the curve y^{2} = 24x meet the curve xy=2 at the points A and B. Then the mid points of such line segments AB lie on a parabola with the

Option: 1

 \small \text Length\; o\!f \;latus \;rectum\;\frac{3}{2}


Option: 2

\small \text \;directrix\; \mathrm{4x} = \mathrm{-3}


Option: 3

\small \text Length\; o\!f \;latus \;rectum\;\mathrm{2}


Option: 4

\small directrix\; \mathrm{4x} = \mathrm{3}


Answers (1)

best_answer

\small \mathrm{c}_1: \mathrm{y}^2=24 x \quad \& \quad \mathrm{c}_2: \mathrm{xy}=2

                           

\small \!\!\!\!\!\!\!\!\!\!\mathrm{AB}: \quad [Tangent\; to\; parabola\; at\; \mathrm{p}(\mathrm{t})]\\ t y=x+6 t^2\; \; \; \; \; \; \; \; \; \; .......(1)\\ \!\!\!\!\!\!A B : \; \; \; [chord\; with\; given\; mid\; point \;o\!f \;hyperbola]\\ \mathrm{T}=\mathrm{S}_1

\small \!\!\!\!\!\!\!\!\!\frac{x}{h}+\frac{y}{k}=2\\ \\ from (1) \& (2)

\small \begin{aligned} & \frac{-1}{\frac{1}{h}}=\frac{\mathrm{t}}{\frac{1}{k}}=\frac{6 \mathrm{t}^2}{2} \\ & -\mathrm{h}=\mathrm{kt}=3 \mathrm{t}^2 \\ & \mathrm{~h}=-3 \mathrm{t}^2 \quad \& \quad k=3 \mathrm{t} \\ & \mathrm{h}=-3 \frac{\mathrm{k}^2}{9} \Rightarrow \mathrm{y}^2=-3 \mathrm{x} \end{aligned}
\small \ell(\mathrm{LR})=3 \quad \& \text { directrix is } \mathrm{x}=\frac{3}{4}

Posted by

rishi.raj

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question