Respuesta :
Given:
The endpoints of the latus rectum at (-2, 3) and (-2, 15).
The directrix at x = 4.
To find:
The equation of the parabola.
Solution:
The equation of the parabola is:
[tex](y-k)^2=4p(x-h)[/tex] ...(1)
Where, [tex]x=h-p[/tex] is directrix and [tex](h+p,k\pm |2p|),p<0[/tex] are the end point of the latus rectum.
The directrix at x = 4. So,
[tex]h-p=4[/tex] ...(i)
The endpoints of the latus rectum at (-2, 3) and (-2, 15). So,
[tex](h+p,k-|2p|)=(-2,3)[/tex]
[tex](h+p,k+|2p|)=(-2,15)[/tex]
Now,
[tex]h+p=-2[/tex] ...(ii)
[tex]k-2p=3[/tex] ...(iii)
[tex]k+2p=15[/tex] ...(iv)
Adding (i) and (ii), we get
[tex]2h=2[/tex]
[tex]h=1[/tex]
Putting [tex]h=1[/tex] in (i), we get
[tex]1-p=4[/tex]
[tex]-p=4-1[/tex]
[tex]-p=3[/tex]
[tex]p=-3[/tex]
Putting [tex]p=-3[/tex] in (iii), we get
[tex]k-|2(-3)|=3[/tex]
[tex]k-6=3[/tex]
[tex]k=3+6[/tex]
[tex]k=9[/tex]
Putting [tex]h=1,p=-3,k=9[/tex] in (1), we get
[tex](y-(9))^2=4(-3)(x-1)[/tex]
[tex](y-9)^2=-12(x-1)[/tex]
Therefore, the required equation of the parabola is [tex](y-9)^2=-12(x-1)[/tex].
