Question #104248
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
1
Expert's answer
2020-03-02T10:31:20-0500

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=2x+y=2

or y=x+2y= -x +2

slope of directrix = -1

since for two perpendicular line having slope m1 and m2, m1m2=1m_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(x3)y-(-4)=1(x-3)

y=x7y=x-7


Axis will met the directrix at point D which coordinate will be found by solving equation y=x+2y=-x+2 and y=x7y=x-7

adding two equation

2y=52y = -5

y=5/2y=-5/2

substituting value in first equation

5/2=x+2-5/2=-x+2

x=9/2x=9/2

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

Coordinate of point V using mid point formula

V=(3+9/22,45/22)=(154,134)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

x2+y22xy8x+20y+c=0x^2+y^2-2xy-8x+20y+c=0

(154)2+(134)22(154)(134)8(154)+20(134)+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

22516+16916+19583065+c=0\frac{225}{16}+\frac{169}{16}+\frac{195}{8}-30-65+c=0

multiply both side by 16

225+169+3904801040+16c=0225+169+390-480-1040+16c=0

736+16c=0-736+16c=0

16c=73616c=736

c=736/16=46c=736/16 = 46

c = 46 is the answer.


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