# What is the equation of the parabola passing through the points (0, 0) and (0,1) and having the line x+y+1=0 as its axis of symmetry?

Jan 25, 2017

Equation of parabola is ${x}^{2} + {y}^{2} + 2 x y + 5 x - y = 0$

#### Explanation:

As axis of symmetry is $x + y + 1 = 0$ and focus lies on it, if abscissa of focus is $p$, ordinate is $- \left(p + 1\right)$ and coordinates of focus are $\left(p , - \left(p + 1\right)\right)$.

Further, directrix will be perpendicular to axis of symmetry and its equation would be of the form $x - y + k = 0$

As every point on parabola is equidistant from focus and directrix, its equation will be

${\left(x - p\right)}^{2} + {\left(y + p + 1\right)}^{2} = {\left(x - y + k\right)}^{2} / 2$

This parabola passes through $\left(0 , 0\right)$ and $\left(0 , 1\right)$ and hence

${p}^{2} + {\left(p + 1\right)}^{2} = {k}^{2} / 2$ .....................(1) and

${p}^{2} + {\left(p + 2\right)}^{2} = {\left(k - 1\right)}^{2} / 2$ .....................(2)

Subtracting (1) from (2), we get

$2 p + 3 = \frac{- 2 k + 1}{2}$, which gives $k = - 2 p - \frac{5}{2}$

This reduces equation of parabola to ${\left(x - p\right)}^{2} + {\left(y + p + 1\right)}^{2} = {\left(x - y - 2 p - \frac{5}{2}\right)}^{2} / 2$

and as it passes through $\left(0 , 0\right)$, we get

${p}^{2} + {p}^{2} + 2 p + 1 = \frac{4 {p}^{2} + 10 p + \frac{25}{4}}{2}$ or $4 p + 2 = \frac{25}{4} + 10 p$

i.e. $6 p = - \frac{17}{4}$ and $p = - \frac{17}{24}$

and hence $k = - 2 \times \left(- \frac{17}{24}\right) - \frac{5}{2} = - \frac{13}{12}$

and equation of parabola as

${\left(x + \frac{17}{24}\right)}^{2} + {\left(y + \frac{7}{24}\right)}^{2} = {\left(x - y - \frac{13}{12}\right)}^{2} / 2$ and multiplying by $576 = {24}^{2}$, we get

or ${\left(24 x + 17\right)}^{2} + {\left(24 y + 7\right)}^{2} = 2 {\left(12 x - 12 y - 13\right)}^{2}$

or 576x^2+816x+289+576y^2+336y+49=2(144x^2+144y^2+169-288xy-312x+312y

or $288 {x}^{2} + 288 {y}^{2} + 576 x y + 1440 x - 288 y = 0$

or ${x}^{2} + {y}^{2} + 2 x y + 5 x - y = 0$
graph{(x^2+y^2+2xy+5x-y)(x+y+1)(12x-12y-13)=0 [-11.42, 8.58, -2.48, 7.52]}