Q83 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
The transformed equation of $3 x^2-6 x y+8 y^2=8$ when the axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction, is
- A$5 x^2+10 x y+17 y^2+16=0$
- Correct answerB$5 x^2+10 x y+17 y^2-16=0$
- C$5 x^2-10 x y+17 y^2-16=0$
- D$5 x^2-10 x y+17 y^2+16=0$
Solution
Given transformed equation is
$$
3 x^2-6 x y+8 y^2=8
$$
Now, $x^{\prime}=x \cos \frac{\pi}{4}-y \sin \frac{\pi}{4}=\frac{x-y}{\sqrt{2}}$
and $y^{\prime}=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}}$
Before transformation the equation is
$$
\begin{aligned}
& 3\left(\frac{x-y}{\sqrt{2}}\right)^2-6\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right)+8\left(\frac{x+y}{\sqrt{2}}\right)^2=8 \\
& \Rightarrow 3\left(x^2+y^2-2 x y\right)-6\left(x^2-y^2\right) \\
& \quad+8\left(x^2+y^2+2 x y\right)=16 \\
& \Rightarrow 5 x^2+17 y^2+10 x y-16=0 .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.