Question

The equation of parabola whose focus is (2, -2) and vertex is (3, 0), is

• A. $y^2 = 24(x-3)$
• B. $4x^2 + y^2 + 4x + 12y - 4xy + 24 = 0$
• C. $4x^2 + y^2 - 4x + 52y - 4xy - 24 = 0$
• D. $y^2 = 48(x-3)$

Answer 1

Answer: (C)

C. $4x^2 + y^2 - 4x + 52y - 4xy - 24 = 0$

Answer 2

To find the equation of the parabola, we use its fundamental definition: a parabola is the locus of a point that is equidistant from a fixed point (the focus) and a fixed line (the directrix).

Given:

• Focus S = (2, -2)
• Vertex V = (3, 0)

1. Find the coordinates of the point F' on the directrix such that V is the midpoint of SF'.

Let F' = $(x', y')$. Since V is the midpoint of SF':

$\left(\frac{2+x'}{2}, \frac{-2+y'}{2}\right) = (3, 0)$

Equating the coordinates:

$\frac{2+x'}{2} = 3 \implies 2+x' = 6 \implies x' = 4$

$\frac{-2+y'}{2} = 0 \implies -2+y' = 0 \implies y' = 2$

So, the point F' on the directrix is (4, 2).

2. Determine the slope of the axis of the parabola.

The axis of the parabola passes through the focus S(2, -2) and the vertex V(3, 0).

Slope of axis ($m_{axis}$) = $\frac{0 - (-2)}{3 - 2} = \frac{2}{1} = 2$.

3. Determine the equation of the directrix.

The directrix is perpendicular to the axis of the parabola.

Slope of directrix ($m_{directrix}$) = $-\frac{1}{m_{axis}} = -\frac{1}{2}$.

The directrix passes through F'(4, 2). Using the point-slope form:

$y - 2 = -\frac{1}{2}(x - 4)$

$2(y - 2) = -(x - 4)$

$2y - 4 = -x + 4$

$x + 2y - 8 = 0$

This is the equation of the directrix.

4. Use the definition of a parabola (PS = PM).

Let P(x, y) be any point on the parabola.

The distance from P to the focus S(2, -2) is PS:

$PS = \sqrt{(x-2)^2 + (y-(-2))^2} = \sqrt{(x-2)^2 + (y+2)^2}$

The distance from P to the directrix ($x + 2y - 8 = 0$) is PM. The formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

$PM = \frac{|x + 2y - 8|}{\sqrt{1^2 + 2^2}} = \frac{|x + 2y - 8|}{\sqrt{5}}$

According to the definition, PS = PM. Squaring both sides:

$PS^2 = PM^2$

$(x-2)^2 + (y+2)^2 = \left(\frac{x + 2y - 8}{\sqrt{5}}\right)^2$

$(x-2)^2 + (y+2)^2 = \frac{(x + 2y - 8)^2}{5}$

$5[(x-2)^2 + (y+2)^2] = (x + 2y - 8)^2$

Expand both sides:

$5[x^2 - 4x + 4 + y^2 + 4y + 4] = x^2 + (2y)^2 + (-8)^2 + 2(x)(2y) + 2(2y)(-8) + 2(x)(-8)$

$5[x^2 + y^2 - 4x + 4y + 8] = x^2 + 4y^2 + 64 + 4xy - 32y - 16x$

$5x^2 + 5y^2 - 20x + 20y + 40 = x^2 + 4y^2 + 4xy - 16x - 32y + 64$

Rearrange all terms to one side to form the general equation of the parabola:

$(5x^2 - x^2) + (5y^2 - 4y^2) - 4xy + (-20x + 16x) + (20y + 32y) + (40 - 64) = 0$

$4x^2 + y^2 - 4xy - 4x + 52y - 24 = 0$

Comparing this equation with the given options, it matches option C.