Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="\mathrm{\text { The vertex of the parabola } 2\left((x-1)^2+(y-2)^2\right)=(x+y+3)^2 \text { is }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B%20The%20

\mathrm{\text { The vertex of the parabola } 2\left((x-1)^2+(y-2)^2\right)=(x+y+3)^2 \text { is }}

Option: 1

\left(-\frac{1}{2},-\frac{1}{2}\right)


Option: 2

\left(-\frac{1}{2},\frac{1}{2}\right)


Option: 3

\left(\frac{1}{2},\frac{1}{2}\right)


Option: 4

\left(\frac{1}{2},-\frac{1}{2}\right)


Answers (1)

best_answer

\mathrm{\begin{aligned} & 2\left((x-1)^2+(y-2)^2\right)=(x+y+3)^2 \\ & \sqrt{(x-1)^2+(y-2)^2}=\frac{|x+y+3|}{\sqrt{2}} \end{aligned}}

Therefore, focus is S(1,2) and directrix is x+y+3=0. Axis of the parabola is x-y+1=0.
Solving directrix and axis, we get foot of perpendicular of directrix on axis as A(-2,-1)
Hence, vertex is mid-point of A S which is \mathrm{\left(-\frac{1}{2}, \frac{1}{2}\right)}.

Posted by

Ritika Kankaria

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 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in
anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in
anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

Latest Question