Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  The eccentricity of the conic represented by <img alt="\mathrm{x^{2}-y^{2}-4 x+4 y+16=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D-y%5E%7B2%7D-4%20x&pl

 The eccentricity of the conic represented by \mathrm{x^{2}-y^{2}-4 x+4 y+16=0} is

Option: 1

1


Option: 2

\sqrt{2}


Option: 3

2


Option: 4

1/2


Answers (1)

best_answer

\text{We have}\: \mathrm{x}^{2}-\mathrm{y}^{2}-4 \mathrm{x}+4 \mathrm{y}+16=0

\text{or,}\: \left(\mathrm{x}^{2}-4 \mathrm{x}\right)-\left(\mathrm{y}^{2}-4 \mathrm{y}\right)=-16

\text{or,}\: \left(x^{2}-4 x+4\right)-\left(y^{2}-4 y+4\right)=-16

\text{or,}\:(x-2)^{2}-(y-2)^{2}=-16$ or, $\frac{(x-2)^{2}}{4^{2}}-\frac{(y-2)^{2}}{4^{2}}=1

Shifting the origin at (2,2), we obtain \mathrm{\frac{X^{2}}{4^{2}}-\frac{Y^{2}}{4^{2}}=-1}, where \mathrm{x=X+2, y=Y=2}

This is a rectangular hyperbola, whose eccentricity is always \mathrm{\sqrt{2}}.

Posted by

Kuldeep Maurya

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

Kota: JEE aspirant dies after falling from 9th floor; police say likely suicide
anam-khan

JEE Main 2026 exam dates won't be postponed; NTA answers students’ top FAQs
anam-khan

JEE Main 2026 application form correction facility from December 1; registration closes on November 27

Latest Question