Q83 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

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 \((\alpha+\beta+\gamma-\sqrt{\delta})^2=\)

  • A
    3
  • Correct answer
    B
    9
  • C
    4
  • D
    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.

Hence, the correct answer is option: 'B'.

Relevant Questions