GATE 2025 IN Q.12: solution and answer

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. The variables separate. Multiply both sides by $ydx$ (with $y\ne 0$) to get $ydy=9xdx$.

2. Integrate both sides: $\int ydy=\int 9xdx\Rightarrow \frac{y^2}{2}=\frac{9x^2}{2}+C_1$.

3. Multiply by 2 and rearrange: $$y^2-9x^2=C,C=2C_1.$$

4. Identify the conic. The squares $x^2$ and $y^2$ have opposite signs, so the curve is a hyperbola. The discriminant test for $A{x}^2+Bxy+C{y}^2+\cdots =0$ gives the same answer: here $A=-9$, $B=0$, $C=1$, so $B^2-4AC=36>0$, which means a hyperbola. For $C>0$ the curve is $\frac{y^2}{C}-\frac{x^2}{C/9}=1$, which opens along the $y$-axis. For $C<0$ it is $\frac{x^2}{|C|/9}-\frac{y^2}{|C|}=1$, which opens along the $x$-axis. In both cases the asymptotes are $y=\pm 3x$. The case $C=0$ is the degenerate hyperbola made of that pair of lines.

5. Rule out the other options. A circle needs equal coefficients with the same sign, as in $x^2+y^2=r^2$. An ellipse needs both squares to have the same sign. A parabola has only one of the variables squared. So the answer is (A), a hyperbola.