Q82 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics
If $x^2=8$ ay is the transformed equation of $x^2-4 y+6 x+$ $15=0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes, then $2 \alpha+8 \beta^2=$
- A8
- B18
- Correct answerC12
- D16
Solution
Translated equation of $x^2-8 a y=0$ is with point $(\alpha$, $\beta$ ) is
$$
\begin{aligned}
& (x-\alpha)^2-8 a(y-\beta) \equiv x^2-4 y+6 x+15 \text { (Given) } \\
& \Rightarrow x^2-8 a y-2 \alpha x+\alpha^2+8 a \beta \\
& =x^2-4 y+6 x+15
\end{aligned}
$$
On comparing we get
$$
\alpha=-3, \beta=\frac{3}{2} \text { and } \mathrm{a}=\frac{1}{2}
$$
Hence $2 \alpha+8 \beta^2=12$
Hence, the correct answer is option: 'C'.