When the axes are rotated through an angle 45°, the transformed equation of a curve is $17X^2-16XY+17Y^2=225$. Find the original equation of the curve.

Given transformed equation of a curve is
$17{\mathrm{X}}^2-16\mathrm{XY}+17{\mathrm{Y}}^2=225 .....\left(1\right)$
$\text{ Given }\mathrm{\theta }={45}^0$
 

$\begin{array}{l}\mathrm{X}=\mathrm{xcos}\mathrm{\theta }+\mathrm{ysin}\mathrm{\theta }=\mathrm{xcos}{45}^0+\mathrm{ysin}{45}^0\\ \mathrm{X}=\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}}\end{array}$
$\begin{array}{l}\mathrm{Y}=-\mathrm{xsin}\mathrm{\theta }+\mathrm{ycos}\mathrm{\theta }=-\mathrm{xsin}{45}^0+\mathrm{ycos}{45}^0\\ =\frac{-\mathrm{x}+\mathrm{y}}{\sqrt{2}}\end{array}$
Substitute X, Y values in eq (1) we get
The original equation is
$\begin{array}{l}17{\left(\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)}^2-16\left(\frac{\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)\left(\frac{-\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)+17{\left(\frac{-\mathrm{x}+\mathrm{y}}{\sqrt{2}}\right)}^2=225\\ \Rightarrow 17\left(\frac{{\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2}{2}\right)-16{\left(\frac{y^2-x^2}{2}\right)}^{ }+17\left(\frac{{\mathrm{y}}^2-2xy+{\mathrm{x}}^2}{2}\right)=225\\ \\ \left.\Rightarrow \frac{17{\mathrm{x}}^2+34\mathrm{xy}+17{\mathrm{y}}^2-16{\mathrm{y}}^2+16{\mathrm{x}}^2+17{\mathrm{y}}^2-34\mathrm{xy}+17{\mathrm{x}}^2}{2}\right)=225\\ \Rightarrow 50{\mathrm{x}}^2+18{\mathrm{y}}^2=450\Rightarrow 2\left(25{\mathrm{x}}^2+9{\mathrm{y}}^2\right)=450\\ \therefore 25{\mathrm{x}}^2+9{\mathrm{y}}^2=225\\ \end{array}$