Search any question & find its solution
Question:
Answered & Verified by Expert
An ellipse has 6 and 2 as length of major and minor axes respectively. If the centre is at $(5,6)$ and the major axis is along $x-y+1=0$, then the ellipse is
Options:
Solution:
1704 Upvotes
Verified Answer
The correct answer is:
$(x+y-11)^2+9(x-y+1)^2=18$
Given, $2 a=6$
$\Rightarrow a=3$ and $2 b=2 \Rightarrow b=1$
Equation of $B C: x+y+k=0$ ...(i)
It passes through $(5,6)$.
$\therefore \quad k=-11$
Equation becomes $x+y-11=0$
Equation of ellipse
$\frac{\left(\frac{x+y-11}{\sqrt{1^2+1^2}}\right)^2}{3^2}+\frac{\left(\frac{x-y+1}{\sqrt{1^2+1^2}}\right)^2}{1^2}=1$
$\begin{aligned} & \Rightarrow \quad \frac{(x+y-11)^2}{18}+\frac{(x-y+1)^2}{2}=1 \\ & \Rightarrow \quad(x+y-11)^2+9(x-y+1)^2=18\end{aligned}$
$\Rightarrow a=3$ and $2 b=2 \Rightarrow b=1$
Equation of $B C: x+y+k=0$ ...(i)
It passes through $(5,6)$.
$\therefore \quad k=-11$
Equation becomes $x+y-11=0$
Equation of ellipse
$\frac{\left(\frac{x+y-11}{\sqrt{1^2+1^2}}\right)^2}{3^2}+\frac{\left(\frac{x-y+1}{\sqrt{1^2+1^2}}\right)^2}{1^2}=1$
$\begin{aligned} & \Rightarrow \quad \frac{(x+y-11)^2}{18}+\frac{(x-y+1)^2}{2}=1 \\ & \Rightarrow \quad(x+y-11)^2+9(x-y+1)^2=18\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.