When the coordinate axes are rotated about the origin in the positive direction through an angle…
When the coordinate axes are rotated about the origin in the positive direction through an angle \(\frac{\pi}{4}\), if the equation \(25 x^2+9 y^2=225\) is transformed to \(\alpha x^2+\beta x y+\gamma y^2=\delta\), then \((\alpha+\beta+\gamma-\sqrt{\delta})^2=\)
3
9
4
16
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
\(\begin{aligned}
& (x, y)=[(X \cos \theta-Y \sin \theta),(Y \cos \theta+X \sin \theta)) \text {, where } \\
& \theta=\frac{\pi}{4} \\
& =\left(\left(\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\right),\left(\frac{Y}{\sqrt{2}}+\frac{X}{\sqrt{2}}\right)\right) \\
\end{aligned}\)
so, \(25 x^2+9 y^2=225\) becomes
\(\begin{aligned}
& 25\left(\frac{X-Y}{\sqrt{2}}\right)^2+9\left(\frac{X+Y}{\sqrt{2}}\right)^2=225 \\
& \Rightarrow 34 X^2+34 Y^2-32 X Y=450 \\
& \Rightarrow 17 X^2+17 Y^2-16 X Y=225
\end{aligned}\)
On comparing, we get
\(\begin{aligned}
& \alpha=\gamma=17, \beta=-16 \text { and } \delta=225 \\
& \therefore(\alpha+\beta+\gamma-\sqrt{\delta})^2 \\
& =(34-16-15)^2 \\
& =3^2=9
\end{aligned}\)
Hence, option (2) is correct.