Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equation to the ellipse whose focus is the point whose directrix is the straight line <img alt=

The equation to the ellipse whose focus is the point (-1,1) whose directrix is the straight line x-y+3=0 and whose eccentricity is \frac{1}{2}, is

Option: 1

\begin{aligned} & 7 x^2+2 x y+7 y^2+10 x-10 y+7=0 \\ \end{aligned}


Option: 2

\begin{aligned} & x^2+2 x y+10 x-10 y+3=0 \\ \end{aligned}


Option: 3

\begin{aligned} & 3 x^2+x y+10 x-10 y+3=0 \end{aligned}


Option: 4

None of the above 


Answers (1)

best_answer

(a) If P(x, y) be any point on the ellipse, S be its focus, and PN be the perpendicular from P on directrix, then by definition of an ellipse P S^2=e^2 P N^2, hence

\begin{aligned} (x+1)^2+(y-1)^2 & =\frac{1}{4}\left(\frac{x-y+3}{\sqrt{2}}\right)^2 \\ \\& =\frac{(x-y+3)^2}{8} \end{aligned}

\text { [as focus is }(-1,1) \text { and directrix is } x-y+3=0]

\begin{aligned} & \Rightarrow \quad 8\left(x^2+y^2+2 x-2 y+2\right) \\ \\& \quad=\quad x^2+y^2+9-2 x y+6 x-6 y \\ \\& \Rightarrow \quad 7 x^2+2 x y+7 y^2+10 x-10 y+7=0 \end{aligned}

Posted by

mansi

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

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question