Question

Find the equation of the parabola whose focus is (5,2) and having vertex at (3,2).

Answer 1

In a parabola,vertex is the mid point of the focus and the point of intersection of the axis and directrix.So let (x,y) be the coordinates of the point of intersection of the axis and directrix.

Then (3,2) is the midpoint of the lines segment joining (5,2) and $x_1,y_1$.

$\frac{x_1+5}{2}=3and\frac{y_1+2}{2}=2x_1+5=6and{y}_1+2=4x_1=1and{y}_1=2$

The directrix meets the axis at(1,2)

Let A be the vertex and S be the focus of the required parabola

Then

$m_1=slopeofAS=\frac{2-2}{5-3}=0$

Let $m_2$ be the slope of the directrix

Then,$m_2=\infty$

$[\because Directrixisperpendiculartotheaxis]$

Thus the directrix passes through (1,2) and the slope $\infty$,so its equation is

$y-2=\infty (x-1)\frac{y-2}{\infty }=x-1x-1=0LetP(x,y)\text{be a point on the parabola}\text{Then PS=distance of P from the directrix}\sqrt{(x-5{)}^2+(y-2{)}^2}=\left|\frac{x-1}{\sqrt{1^2}}\right|(x-5{)}^2+(y-2{)}^2=(x-1{)}^2{x}^3+25-10x+y^2+4-4y=x^2+1-2x{y}^2-4y-8x+28=0\text{Hence the required of the parabola is}{y}^2-4y-8x+28=0.$