Q82 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$(a, b)$ is the point to which the origin has to be shifted by translation of axes so as to remove the first-degree terms from the equation $2 x^2-3 x y+4 y^2+5 y-6=0$. If the angle by which the axes are to be rotated in positive direction about the origin to remove the $x y$-term from the equation $a x^2+23 a b x y+b y^2=0$ is $\theta$, then $\tan 2 \theta=$
- A$\frac{\pi}{4}$
- Correct answerB$60$
- C$\frac{\pi}{3}$
- D$15$
Solution
Shifting the origin by translation
$\left.\begin{array}{l}a=x-h ; b=x-k \\ \Rightarrow x=a+h ; y=b+k\end{array}\right\}$ ...(i)
Substituting in equation $2 x^2-3 x y+4 y^2+5 y-6=0$
$\begin{aligned}
& 2(a+h)^2-3(a+h)(b+k)+4(b+k)^2+5(b+k)-6=0 \\
& \Rightarrow 2 a^2+4 b^2-3 a b+(4 h-3 k) a+(8 k-3 h+5) b-6=0
\end{aligned}$
To remove first degree term, $\begin{aligned} & 4 h-3 k=0 \text { and } 8 k-3 h+5=0 \\ & \Rightarrow h=\frac{-15}{23}, k=\frac{-20}{23} \end{aligned}$ Since at origin $(x, y)=(0,0) \therefore$ From (i), $a=\frac{15}{23}, b=\frac{20}{23}$ Let $\left(x^{\prime}, y^{\prime}\right)$ be point after rotation of axes along $\theta$. $x^{\prime} \cos \theta+y^{\prime} \sin \theta=x ; x^{\prime} \sin \theta-y^{\prime} \cos \theta=-y$ Now we have to remove $x y$-term from equation $a x^2+23 a b x y+b y^2=0$ by rotation along $\theta$. Substituting $(a, b)$ we get, $\frac{15}{23} x^2+\frac{(15)(20)}{23} x y+\frac{20}{23} y^2=0 \Rightarrow 3 x^2+60 x y+4 y^2=0$
Substituting $(x, y)$ we get, $\begin{aligned} & 3\left(x^{\prime} \cos \theta+y^{\prime} \sin \theta\right)^2+60\left(x^{\prime} \cos \theta+y^{\prime} \sin \theta\right) \left(y^{\prime} \cos \theta-x^{\prime} \sin \theta\right)+4\left(x^{\prime} \sin \theta-y^{\prime} \cos \theta\right)^2=0 \\ & \Rightarrow 3\left(x^{\prime 2} \cos ^2 \theta+y^{\prime 2} \sin ^2 \theta+2 x^{\prime} y^{\prime} \cos \theta \sin \theta\right) +60\left(-x^{\prime 2} \cos \theta \sin \theta-x^{\prime} y^{\prime} \sin ^2 \theta+x^{\prime} y^{\prime} \cos ^2 \theta +y^{\prime 2} \sin \theta \cos \theta\right)+4\left(y^{\prime 2} \cos ^2 \theta+x^{\prime 2} \sin ^2 \theta-2 x^{\prime} y^{\prime} \sin \theta \cos \theta\right)=0 \end{aligned}$
To Remove $x^{\prime} y^{\prime}$-term Co-efficient of $x^{\prime} y^{\prime}$-term must be 0 $\begin{aligned} & \therefore 6 \cos \theta \sin \theta-60 \sin ^2 \theta+60 \cos ^2 \theta-8 \sin \theta \cos \theta=0 \\ & \Rightarrow 60\left(\cos ^2 \theta-\sin ^2 \theta\right)-2 \sin \theta \cos \theta=0 \\ & \Rightarrow 60 \cos 2 \theta-\sin 2 \theta=0 \Rightarrow 60 \cos 2 \theta=\sin 2 \theta \\ & \Rightarrow 60=\tan 2 \theta \end{aligned}$
To remove first degree term, $\begin{aligned} & 4 h-3 k=0 \text { and } 8 k-3 h+5=0 \\ & \Rightarrow h=\frac{-15}{23}, k=\frac{-20}{23} \end{aligned}$ Since at origin $(x, y)=(0,0) \therefore$ From (i), $a=\frac{15}{23}, b=\frac{20}{23}$ Let $\left(x^{\prime}, y^{\prime}\right)$ be point after rotation of axes along $\theta$. $x^{\prime} \cos \theta+y^{\prime} \sin \theta=x ; x^{\prime} \sin \theta-y^{\prime} \cos \theta=-y$ Now we have to remove $x y$-term from equation $a x^2+23 a b x y+b y^2=0$ by rotation along $\theta$. Substituting $(a, b)$ we get, $\frac{15}{23} x^2+\frac{(15)(20)}{23} x y+\frac{20}{23} y^2=0 \Rightarrow 3 x^2+60 x y+4 y^2=0$
Substituting $(x, y)$ we get, $\begin{aligned} & 3\left(x^{\prime} \cos \theta+y^{\prime} \sin \theta\right)^2+60\left(x^{\prime} \cos \theta+y^{\prime} \sin \theta\right) \left(y^{\prime} \cos \theta-x^{\prime} \sin \theta\right)+4\left(x^{\prime} \sin \theta-y^{\prime} \cos \theta\right)^2=0 \\ & \Rightarrow 3\left(x^{\prime 2} \cos ^2 \theta+y^{\prime 2} \sin ^2 \theta+2 x^{\prime} y^{\prime} \cos \theta \sin \theta\right) +60\left(-x^{\prime 2} \cos \theta \sin \theta-x^{\prime} y^{\prime} \sin ^2 \theta+x^{\prime} y^{\prime} \cos ^2 \theta +y^{\prime 2} \sin \theta \cos \theta\right)+4\left(y^{\prime 2} \cos ^2 \theta+x^{\prime 2} \sin ^2 \theta-2 x^{\prime} y^{\prime} \sin \theta \cos \theta\right)=0 \end{aligned}$
To Remove $x^{\prime} y^{\prime}$-term Co-efficient of $x^{\prime} y^{\prime}$-term must be 0 $\begin{aligned} & \therefore 6 \cos \theta \sin \theta-60 \sin ^2 \theta+60 \cos ^2 \theta-8 \sin \theta \cos \theta=0 \\ & \Rightarrow 60\left(\cos ^2 \theta-\sin ^2 \theta\right)-2 \sin \theta \cos \theta=0 \\ & \Rightarrow 60 \cos 2 \theta-\sin 2 \theta=0 \Rightarrow 60 \cos 2 \theta=\sin 2 \theta \\ & \Rightarrow 60=\tan 2 \theta \end{aligned}$
Hence, the correct answer is option: 'B'.