Question

The equation of reflection of the ellipse $\frac{(x-4{)}^2}{16}+\frac{(y-3{)}^2}{9}=1$ about the line $x-y-2=0$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $\frac{(x-2{)}^2}{16}+\frac{(y-5{)}^2}{9}=1$
• B. $\frac{(x-2{)}^2}{9}+\frac{(y-5{)}^2}{16}=1$
• C. $\frac{(x-5{)}^2}{16}+\frac{(y-2{)}^2}{9}=1$
• D. $\frac{(x-5{)}^2}{9}+\frac{(y-2{)}^2}{16}=1$

Answer 1

The correct option is D $\frac{(x-5{)}^2}{9}+\frac{(y-2{)}^2}{16}=1$

Given : $\frac{(x-4{)}^2}{16}+\frac{(y-3{)}^2}{9}=1$

Any point $P$ on the curve is
$P=\left(4\right(1+\cos \theta ),3(1+\sin \theta \left)\right)$
Let the reflection of $P$ be $(h,k)$

Now, the reflection of the point $P$ about the line $x-y-2=0$ is $\frac{h-4(1+\cos \theta )}{1}=\frac{k-3(1+\sin \theta )}{-1}=\frac{-2\left[4\right(1+\cos \theta )-3(1+\sin \theta )-2]}{2}=1-4\cos \theta +3\sin \theta$

So,
$\Rightarrow h=5+3\sin \theta ,k=2+4\cos \theta \Rightarrow {\left(\frac{h-5}{3}\right)}^2+{\left(\frac{k-2}{4}\right)}^2=1$

Hence, the equation of reflected ellipse is
$\frac{(x-5{)}^2}{9}+\frac{(y-2{)}^2}{16}=1$