Question

Prove that the equation of the parabola whose vertex and focus are on the $x-$axis at a distance $a$ and $a^{\prime }$ from the origin respectively is $y^2=4(a^{\prime }-a)(x-a)$.

Answer 1

According to the question,

First, We have to find the slope $m$ of the axis of symmetry.

$m=\frac{1-(-1)}{2-1}=2$

the directrix, being perpendicular to the axis of symmetry will have a slope of $-\frac{1}{2}$

we know that the directrix passes through the point $(3,3)$, and so using the point-slope formula, we can find the equation of the directrix as follow,

$y-3=-\frac{1}{2}(x-3)$

Arranging this in slope-intercept form, we obtain,

$y=-\frac{1}{2}x+\frac{9}{2}$

Now the parabola we are seeking will be the locus of points equidistant from the directrix and the focus, And so using the formulas for the distance between two points and between a point and a line,

$(x-1{)}^2+(y+1{)}^2=\frac{{(-\frac{1}{2}x+\frac{9}{2}-y)}^2}{{(-\frac{1}{2})}^2+1}$

$(x-1{)}^2+(y+1{)}^2=\frac{{(9-x-2y)}^2}{5}$

$5x^2-10x+5+5y^2+10y+5=x^2+4xy-18x+4y^2-36y+81$

$4x^2+y^2-4xy+8x+46y-71=0$

Then the equation of parabola will be

$4x^2+y^2-4xy+8x+46y-71=0$