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Suppose the equation of latus rectum is x=4 and the vertex is (2,3). I am confused wouldn't there be many parabola with this same vertex and latus rectum.If not how to find the equation?

The answer that is given is $$(y-3)^2=8(x-2)$$ in the textbook. I am not getting it. Thanks in advance.

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  • $\begingroup$ First: what is the distance of your given point from your latus rectum? $\endgroup$ – J. M. is a poor mathematician Apr 9 '13 at 10:20
  • $\begingroup$ The distance is not given.I need to find the focus I guess to gauge the distance. This is all the imformation that is given. $\endgroup$ – Neer Apr 9 '13 at 10:22
  • $\begingroup$ It isn't given, but you can determine the distance of your vertex from the latus rectum. Draw a picture and see. But, I suppose this is all moot, as someone else has seen it fit to spoonfeed you... $\endgroup$ – J. M. is a poor mathematician Apr 9 '13 at 10:24
  • $\begingroup$ No I can find the distance from the focus . Its coming 2 but the ordinate can be something else so the distance is also changing. $\endgroup$ – Neer Apr 9 '13 at 10:29
  • $\begingroup$ Nono, forget about the focus for now! I was not asking for it! Again: what is the distance from the line you were given to the vertex you were given? $\endgroup$ – J. M. is a poor mathematician Apr 9 '13 at 10:30
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enter image description here

Let $BE$ be the directrix

As the axis of the parabola is perpendicular to the latus rectum, the equation of the axis(OA) $y=3$

So, $c=a=3$

As the vertex O$(2,3)$ is the mid-point of $A,B, \frac{b+4}2=2\implies b=0$
So, the equation of the directrix(BE) will be $x=0$ as it is perpendicular to the axis.

If P$(h,k)$ be any point on the parabola,

the distance of $AP=\sqrt{(h-4)^2+(k-3)^2}$

the distance of P$(h,k)$ from the directrix $BE$ will be $|h|$

As the eccentricity of a parabola is $1$

So, $$\sqrt{(h-4)^2+(k-3)^2}=|h|$$

Squaring we get $(h-4)^2+(k-3)^2=h^2\implies (k-3)^2=h^2-(h-4)^2=8h-16=8(h-2)$

So, the equation of the parabola : $(y-3)^2=8(x-2)$

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  • $\begingroup$ How did you get the "a"? Please specify.I found the focus too but the two ends of latus rectum is given (4,7) (4,1) in the answer so that makes focus (4,4) I guess. Where am I missing? $\endgroup$ – Neer Apr 9 '13 at 10:27
  • $\begingroup$ @Neer, I'm going to insert the image with explanation $\endgroup$ – lab bhattacharjee Apr 9 '13 at 10:31
  • $\begingroup$ Hey thanks for your help. No needd to upload the image. I got it. Maybe the two intersection's coordinate will be (4,7) and (4,-1). That - was not printed that caused all the problems0 Thanks for that image. And that "a" question was very stupid from my side. $\endgroup$ – Neer Apr 9 '13 at 10:59

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