Q84 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

The equation of a curve $C$ is transformed to $X^2+Y^2-6 X+8 Y+21=0$ by the rotation of coordinate axes about the origin through an angle of $\frac{\pi}{4}$ in the positive direction of $X$-axis. If $a x^2+b y^2+c x+d y+e=0$ is the equation of the curve $C$ before the transformation, then $\left(a+b+c^2+d^2-5 e\right)^2=0$.

  • A
    4
  • Correct answer
    B
    9
  • C
    16
  • D
    25

Solution

Given, transformed equation is $x^2+y^2-6 x+8 y+21=0$ Now, $\begin{aligned} & X^{\prime}=x \cos \pi / 4-Y \sin \pi / 4=\frac{X-Y}{\sqrt{2}} \\ & Y^{\prime}=x \sin \pi / 4+Y \cos \pi / 4=\frac{X+Y}{2} \end{aligned}$ Before transformation, the equation is $\begin{aligned} & \left(\frac{X-Y}{\sqrt{2}}\right)^2+\left(\frac{X+Y}{\sqrt{2}}\right)^2-6\left(\frac{X-Y}{\sqrt{2}}\right)+8\left(\frac{X+Y}{\sqrt{2}}\right) \\ & \Rightarrow \quad X^2+Y^2-2 X Y+X^2+Y^2+2 X Y \quad 0 \\ & \quad-6 \sqrt{2} X+6 \sqrt{2} Y+8 \sqrt{2} X+8 \sqrt{2} Y+42=0 \\ & =2 x^2+2 y^2+2 \sqrt{2} x+14 \sqrt{2} y+42=0 \\ & \Rightarrow \quad X^2+Y^2+\sqrt{2} X+7 \sqrt{2} Y+21=0 \end{aligned}$ comparing this to, $\begin{aligned} & a X^2+b Y^2+c X+d Y+e=0, \text { we get } \\ & a=1, b=1, c=\sqrt{2}, d=7 \sqrt{2}, e=21 \\ \therefore \quad & \left(a+b+c^2+d^2-5 e\right)^2 \\ & =(1+1+2+98-105)^2 \\ & =(102-105)^2 \\ & =9 \end{aligned}$

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

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