Question

For the given parabola $9y^2-16x-12y-57=0$, which of the following is/are correct?

• A. axis will be $y=2$
• B. axis will be $3y=2$
• C. vertex will be $(61,2)$
• D. vertex will be $(-\frac{61}{16},\frac{2}{3})$

Answer 1

Answer: (B), (D)

The correct options are
B axis will be $3y=2$
D vertex will be $(-\frac{61}{16},\frac{2}{3})$

We have, $9y^2-16x-12y-57=0$
$\Rightarrow y^2-\frac{4}{3}y=\frac{16}{9}x+\frac{57}{9}\Rightarrow {(y-\frac{2}{3})}^2=\frac{16}{9}x+\frac{57}{9}+\frac{4}{9}=\frac{16}{9}(x+\frac{61}{16})$
Hence vertex will be $(-\frac{61}{16},\frac{2}{3})$
Shifting the origin at $(-\frac{61}{16},\frac{2}{3})$
We have $x=X-\frac{61}{16},y=Y+\frac{2}{3}$
The given equation reduced to $Y^2=\frac{16}{9}X$ which is a parabola with axis $Y=0\Rightarrow 3y=2$