Answer to Question #104248 in Analytic Geometry for Deepak Rana

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

Concept:

For a parabola, axis of parabola is perpendicular to the directrix. we also know distance from parabola to its focus and directix are equal which means DV= VF i.e. V is the mid point of the perpendicular line connecting focus F and to the directrix at D .

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Let focus (F) = (3, -4)

Since equation of directrix is known

$x+y=2$

or $y= -x +2$

slope of directrix = -1

since for two perpendicular line having slope m1 and m2, $m_1*m_2=-1$

slope of axis of parabola = -1/(slope of directrix) =-1/(-1) = 1

equation of axis of parabola will be line having slope =1 and passing through focus F=(3, -4)

$y-(-4)=1(x-3)$

$y=x-7$

Axis will met the directrix at point D which coordinate will be found by solving equation $y=-x+2$ and $y=x-7$

adding two equation

$2y = -5$

$y=-5/2$

substituting value in first equation

$-5/2=-x+2$

$x=9/2$

Thus point D = ( 9/2 , -5/2)

Coordinate of point V using mid point formula

$V=(\frac{3+9/2}{2}, \frac{-4-5/2}{2})=(\frac{15}{4}, -\frac{13}{4})$

Since V lies on parabola, it will satisfy the equation of parabola

$x^2+y^2-2xy-8x+20y+c=0$

$\left(\frac{15}{4}\right)^2+\left(-\frac{13}{4}\right)^2-2\left(\frac{15}{4}\right)\left(-\frac{13}{4}\right)-8\left(\frac{15}{4}\right)+20\left(-\frac{13}{4}\right)+c=0$

$\frac{225}{16}+\frac{169}{16}+\frac{195}{8}-30-65+c=0$

multiply both side by 16

$225+169+390-480-1040+16c=0$

$-736+16c=0$

$16c=736$

$c=736/16 = 46$

c = 46 is the answer.