Question

The equation of the parabola whose focus is (3,5), vertex is (1, 3) is

• A. $x^2 - 2xy + y^2 - 12x - 20y + 68 = 0$
• B. $y^2 + 2xy - y^2 - 12x - 20y = 0$

Answer 1

Answer: (A)

A) $x^2 - 2xy + y^2 - 12x - 20y + 68 = 0$

Answer 2

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

Given:
Focus S = (3, 5)
Vertex V = (1, 3)

1. Find the axis of the parabola:
The axis of the parabola is the line passing through the focus and the vertex.
Slope of the axis ($m_{axis}$) = $\frac{5 - 3}{3 - 1} = \frac{2}{2} = 1$.
Equation of the axis: $y - y_1 = m_{axis}(x - x_1)$
Using vertex V(1, 3): $y - 3 = 1(x - 1)$
$y - 3 = x - 1$
$x - y + 2 = 0$

2. Find the distance 'a' (distance from vertex to focus):
$a = \text{VS} = \sqrt{(3-1)^2 + (5-3)^2}$
$a = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

3. Find the equation of the directrix:
The directrix is perpendicular to the axis of the parabola and is at a distance 'a' from the vertex, on the opposite side of the focus.
Since the slope of the axis is 1, the slope of the directrix ($m_{directrix}$) is $-1$ (negative reciprocal).
Let the equation of the directrix be $y = -x + c$, or $x + y - c = 0$.

Alternatively, let F' be the point on the directrix such that V is the midpoint of SF'.
If F' = $(x', y')$, then V = $\left(\frac{3+x'}{2}, \frac{5+y'}{2}\right)$.
Given V = (1, 3):
$\frac{3+x'}{2} = 1 \implies 3+x' = 2 \implies x' = -1$.
$\frac{5+y'}{2} = 3 \implies 5+y' = 6 \implies y' = 1$.
So, the directrix passes through F'(-1, 1).
Using the point-slope form for the directrix:
$y - 1 = -1(x - (-1))$
$y - 1 = -1(x + 1)$
$y - 1 = -x - 1$
$x + y = 0$

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(3, 5) is PS.
$PS = \sqrt{(x-3)^2 + (y-5)^2}$

The distance from P to the directrix ($x + y = 0$) is PM.
$PM = \frac{|x + y|}{\sqrt{1^2 + 1^2}} = \frac{|x + y|}{\sqrt{2}}$

According to the definition of a parabola, PS = PM.
$\sqrt{(x-3)^2 + (y-5)^2} = \frac{|x + y|}{\sqrt{2}}$

Square both sides to eliminate the square root and absolute value:
$(x-3)^2 + (y-5)^2 = \frac{(x + y)^2}{2}$
$2[(x-3)^2 + (y-5)^2] = (x + y)^2$
Expand the terms:
$2[x^2 - 6x + 9 + y^2 - 10y + 25] = x^2 + 2xy + y^2$
$2[x^2 + y^2 - 6x - 10y + 34] = x^2 + 2xy + y^2$
$2x^2 + 2y^2 - 12x - 20y + 68 = x^2 + 2xy + y^2$

Rearrange the terms to form the general equation of the parabola:
$2x^2 - x^2 + 2y^2 - y^2 - 2xy - 12x - 20y + 68 = 0$
$x^2 + y^2 - 2xy - 12x - 20y + 68 = 0$

Comparing this with the given options, it matches option A.