Question

If the equation of a parabola with the focus at (3,−4) and the directrix x+y = 2 is x^2 +y^2 −2xy−8x+20y+c = 0, then what is the value of c?

Answer 1

We have given that focus of the parabola is $A=$ $(3,-4)$ , equation of directrix is $x+y=2$

,and equation of the parabola is $x^2+y^2-2xy-8x+20y+c=0$ , now we have to find the value of $c$ .If we plot the equation of the parabola, clearly, the parabola will looks like the below figure,

We know that the principle axis of the parabola passes through the focus and perpendicular to it's directrix, Suppose the slope of the directrix is $m_1$ and slope of the principle axis  is $m_2$ , thus $m_1=-1$ .As directrix and principle axis perpendicular to each other, hence $m_1 \cdot m_2 =-1 \implies m_2=1$ ,thus equation principle axis of the parabola is $y=m_2x+c_1 \implies y=x+c_1$ ,also note that principle axis passes through focus , therefore $A=(3,-4)$ will satisfy the equation of principle axis, which implies $-4=3+c_1 \implies c_1 = -7$ . thus equation of principle axis is $y=x-7$ .

Now, $x+y=3 \& y=x-7$ intersect each other let it be at $B$ , hence on solving these two equation we get the intersection point is $B=$ $(\frac{9}{2} ,\frac{-5}{2})$ . We also know that vertex of the parabola $C$ divides line segment $\overline{AB}$ in $1:1$ ratio, moreover $C$ is the middle point of $\overline{AB}$ .let's $C=(x_1,y_1)$ and we know that if $(x_2,y_2) \: \& (x_3,y_3)$ are two point of any line segment then middle point of that line segment is given by,

$(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$

Thus, from above formula, we get,

$C=(x_1,y_1)=(\frac{3+ \frac{9}{2}}{2},\frac{-4 -\frac{5}{2}}{2})=(\frac{15}{4},-\frac{13}{4})$ and clearly, $C$ lies on the parabola, hence it will satisfy the equation of the parabola $x^2+y^2-2xy-8x+20y+c=0 \implies (\frac{15}{4})^2+(\frac{-13}{4})^2-$

$2(\frac{15}{4})(-\frac{13}{4}) - 8(\frac{15}{4}) +20(-\frac{13}{4}) +c=0 \implies c-46=0 \implies c=46.$

Hence, we are done.