Question

The equation of the parabola whose focus $(3,2)$ and vertex $(1,2)$, is

• A. $x^2+4x-8y+12=0$
• B. $x^2-4x-8y+12=0$
• C. $y^2-8x-4y+12=0$
• D. $y^2+4y-8x+12=0$

Answer 1

Answer: (C)

Let $(x_1,y_1)$ be the coordinates of the point of intersection of the axis and the directrix. Then the vertex is the mid point of the line segment joining $(x_1,y_1)$ and focus $(3,2)$
$\Rightarrow \frac{x_1+3}{2}=1$
$\Rightarrow x_1+3=2$
$\Rightarrow x_1=-1$
and $\frac{y_1+2}{2}=2$
$\Rightarrow y_1+2=4$
$\Rightarrow y_1=2$
$\Rightarrow$ The directrix meets the axis at $\left(-1,2\right)$
Let $m_1$ be the slope of the axis.
Then $m_1$= slope of the line joining the focus and the vertex $=\frac{2-2}{3-1}=\frac{0}{2}$
$\therefore$ Directrix passes through $\left(-1,2\right)$ and have
slope $\frac{-2}{0}\therefore x=-1$ is the directrix.
$\therefore$ By definition, equation of parabola is
$\sqrt{{\left(x-3\right)}^2+{\left(y-2\right)}^2}=\left(x+1\right)$
$\Rightarrow {\left(x-3\right)}^2+{\left(y-2\right)}^2={\left(x+1\right)}^2$
$\Rightarrow y^2-4y-8x+12=0$