The equation of the latus rectum of a parabola is given by y=-3. The axis of the parabol is x=0 and vertex (0,0). Find the length of the focal chord that meets the parabola at (2,-1/3)   ans: 100/3

aruv | Student

Latus rectum is focal chord which is perpendicular to the axis of the parabola.

Thus focus of the parabola =(0,-3)

Equation of directrix of parabola is  y=3

vertex (0,0)

Let P(x,y) be point on the parabola then by def of parabola

`sqrt((x-0)^2+(y+3)^2)=sqrt((y-3)^2)`

`` squaring both side and simplify

`x^2=(y-3)^2-(y+3)^2`

`x^2=-12y`                            (i)

equation of focal chord posses through (2,-1/3)

`y+3=(-1/3+3)/(2-0)(x-0)`

`y+3=-(8/6)x`

`y=-3-(8x)/6`                         (ii)

(i) and (ii) will intersect each other at point say Q

`x^2=-12(-3-(8x)/6)`

`x^2=2(18+8x)`

`x^2-16x-36=0`

`x^2-18x+2x-36=0`

`(x-18)+2(x+2)=0`

`x=18,-2`

when  x=18 then

`y=-3-(8xx18)/6`

`y=-3-24=-27`

Thus coordinate of Q( 18,-27)

Thus distance between Q(18,-27) and (2,-1/3)

`d=sqrt((18-2)^2+(-27+1/3)^2)`

`d=sqrt(16^2+(-80/3)^2)`

`d=(1/3)sqrt( 16^2xx3^2+80^2)=93.29/3`

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