Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi}{4}$, if the equation $25 x^2 + 9y^2 = 225$ is transformed to $\alpha x^2 + \beta xy + \gamma y^2 = \delta$, then $(\alpha + \beta + \gamma - \sqrt{\delta})^2$ =

AP EAMCETAP EAMCET 2019

Solution:

After rotation of coordinate axes about the origin in the positive direction through on angle $\frac{\pi}{4}$, the new coordinates are $(X, Y)$ have relation with older coordinates $(x, y)$ is
$(x, y)=[(X \cos \theta-Y \sin \theta),(Y \cos \theta+X \sin \theta))$, where
$\theta=\frac{\pi}{4}$
$=\left(\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\right),\left(\frac{Y}{\sqrt{2}}+\frac{X}{\sqrt{2}}\right)\right)$
so, $25 x^{2}+9 y^{2}=225$ becomes
$25\left(\frac{X-Y}{\sqrt{2}}\right)^{2}+9\left(\frac{X+Y}{\sqrt{2}}\right)^{2}=225$
$\Rightarrow 34 X^{2}+34 Y^{2}-32 X Y=450$
$\Rightarrow 17 X^{2}+17 Y^{2}-16 X Y=225$
On comparing, we get
$\alpha=\gamma=17, \beta=-16$ and $\delta=225$
$\therefore (\alpha+\beta+\gamma-\sqrt{\delta})^{2}$
$=(34-16-15)^{2}$
$=3^{2}=9$