Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let  be a parabola with vertex <img alt="\mathrm{(3,2)}" src="https://entrancecorner.

Let \mathrm{P}_{1} be a parabola with vertex \mathrm{(3,2)} and focus \mathrm{(4,4)} and \mathrm{{P}_{2}} be its mirror image with respect to the line \mathrm{x+2 y=6}. Then the directrix of \mathrm{P_{2}\; is \;x+2 y=}_____

Option: 1

10


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

For P_1:
\\\mathrm{\text{slope of axis}=\frac{4-2}{4-3}=2} \\ \text{Distance between vertex and focus: }\mathrm{\sqrt{5}}\\ \text{Now line is perpendicular to axis}

\text{focus of }\mathrm{P_2: \frac{x-4}{1}=\frac{y-4}{2}=\frac{-2\left ( 4+8-6 \right )}{5}}
                                        \mathrm{x= \frac{-12}{5}+4=\frac{8}{5}

                                       \mathrm{y= \frac{-24}{5}+4=\frac{-4}{5}

\text{Let directrix of } \mathrm{P_2:x+2y=\lambda }

    \begin{aligned} &\therefore\left|\frac{\frac{8}{5}-\frac{8}{5}-\lambda}{\sqrt{5}}\right|=2 \sqrt{5}\\ &|-\lambda|=10\\ &\lambda=\pm 10 \end{aligned}

\mathrm{\lambda =-10\: }\text{is rejected because directrix can't cut parabola}

\mathrm{\therefore \lambda =10} is the answer.

Posted by

admin

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

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

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question