Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A point is taken on the director circle of the circle \mathrm{x^{2}+y^{2}=8} and tangents are drawn from it to the parabola \mathrm{y^{2}=16 x}.the locus of the middle points of chord of contact will be

Option: 1

\mathrm{y^{2}+\left(y^{2}-8 x\right)^{2}=1024}


Option: 2

64 \mathrm{y}^{2}+\left(\mathrm{y}^{2}-8 \mathrm{x}\right)^{2}=1024


Option: 3

\mathrm{y}^{2}-\left(\mathrm{y}^{2}-8 \mathrm{x}\right)^{2}=1024


Option: 4

64 \mathrm{y}^{2}-\left(\mathrm{y}^{2}-8 \mathrm{x}\right)^{2}=1024


Answers (1)

best_answer

Director circle of the given circle \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ will be $\mathrm{x}^{2}+\mathrm{y}^{2}=(a \sqrt{2})^{2}
\mathrm{\Rightarrow \quad \mathrm{x}^{2}+\mathrm{y}^{2}=16}
now point on it will be \mathrm{(4 \cos \theta, 4 \sin \theta)}

Now chord of contact for the parabola \mathrm{\mathrm{y}^{2}=16 \mathrm{x}, \mathrm{T}=0}
\mathrm{\mathrm{y} \times 4 \sin \theta=8(\mathrm{x}+4 \cos \theta)}\quad \ldots(i)

equation of the chord whose middle point is \mathrm{(\mathrm{h}, \mathrm{k})}.

\mathrm{\mathrm{T}=\mathrm{S}_{1}}
\mathrm{\mathrm{ky}-8(\mathrm{x}+\mathrm{h})=\mathrm{k}^{2}-16 \mathrm{~h}}
\mathrm{8 \mathrm{x}-\mathrm{ky}+\mathrm{k}^{2}-8 \mathrm{~h}=0}\quad \ldots(ii)

comparing the coefficient of the two equation of the same line
\mathrm{\frac{k}{4}=\sin \theta, \quad \frac{k^{2}-8 h}{32}=\cos \theta}
\mathrm{so \: 64 y^{2}+\left(y^{2}-8 x\right)^{2}=1024}
 

Posted by

Rakesh

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 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation
anam-khan

Four Kashmiri girls defy the odds, score 99+ percentile in JEE Main 2025
anam-khan

JEE, NEET, CUET coaching centres: Assam, Rajasthan have framed laws, but will they work?

Latest Question