Q98 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
The equation of the pair of asymptotes of the hyperbola $\frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1$ is
- A$2 x^2-3 y^2-12 x+12 y-6=0$
- B$2 x^2-3 y^2-12 x+12 y+8=0$
- C$2 x^2-3 y^2-12 x+12 y-8=0$
- Correct answerD$2 x^2-3 y^2-12 x+12 y+6=0$
Solution
Equation of hyperbola is
$$
\begin{aligned}
& \frac{(x-3)^2}{3}-\frac{(y-2)^2}{2}=1 \\
& \Rightarrow 2\left(x^2+9-6 x\right)-3\left(y^2+4-4 y\right)=6 \\
& \Rightarrow 2 x^2-3 y^2-12 x+12 y+6=6 \\
& \Rightarrow 2 x^2-3 y^2-12 x+12 y=0
\end{aligned}
$$
Pair of asymptotes of the curve differs by the constant only.
$\therefore$ Equation of pair of asymptotes is :
$$
2 x^2-3 y^2-12 x+12 y+\lambda=0
$$
which represents a pair of straight line if
$$
\Delta=a b c+2 f g h-a f^2-b g^2-c h^2=0
$$
where $\mathrm{a}=2, \mathrm{~b}=-3, \mathrm{~h}=0, \mathrm{~g}=-6$
$$
\mathrm{f}=6 \mathrm{c}=\lambda
$$
Then, we have :
$$
\begin{aligned}
& 2 \times(-3) \lambda+2 \times 6 \times(-6) \times 0-2(+6)^2-(-3)(-6)^2-\lambda \\
& \times 0=0 \Rightarrow \lambda=6
\end{aligned}
$$
Putting the value of $\lambda$ in eq ${ }^{\mathrm{n}}$ (i), we get
$$
2 x^2-3 y^2-12 x+12 y+6=0
$$
Hence, the correct answer is option: 'D'.