Video Transcript
Find the equation of a parabola with a focus of negative one, negative three and a directrix of π¦ equals negative five. Give your answer in the form π¦ equals ππ₯ squared plus ππ₯ plus π.
To be able to solve this problem, the first thing we need to look at is what are a focus and a directrix. Well, the focus and the directrix are a point and a line that every point on our parabola is in equal distance from. So to demonstrate this, Iβve actually drawn a couple of lines on the actual sketch. What weβve got is we can see a point, where the distance between the directrix and the point on the parabola is π₯ and therefore the distance between the parabola and the focus is also π₯.
And again, weβve got another point. Iβve called the distance between the focus and the parabola π¦. And therefore, the distance between the parabola and directrix is gonna be π¦. And itβs this relationship that we can actually use to help us solve the problem. So first of all, what Iβm gonna do is pick a point on my parabola. So Iβm gonna call it π₯, π¦.
And first of all, what I want to do is actually want to work out the distance between our point π₯, π¦ and the focus at negative one, negative three. And to do that, what Iβm gonna use is the distance formula. And the distance formula tells us that the distance between two points is equal to the square root of π₯ two minus π₯ one all squared plus π¦ two minus π¦ one all squared. So weβre gonna look at our points. Weβve got π₯, π¦ and negative one, negative three.
And then what Iβve done is Iβve actually labelled our points. Iβve actually called them π₯ two, π¦ two and π₯ one, π¦ one. Iβve done it this way around just because itβll be easier when we actually look to simplify later on. So therefore, we can say that the distance is equal to the square root of π₯ plus one all squared. And thatβs because our π₯ two is π₯ and our π₯ one is negative one. And if you subtract a negative, it turns positive. And then, this is plus: π¦ plus three all squared.
Okay, great, so weβve now got the distance between the focus and our point π₯, π¦. Weβre now gonna look at the distance from our point and our directrix, which is π¦ equals negative five. And as we always have a vertical line from our directrix to our point on our parabola, then we donβt actually have to worry about the π₯-coordinates because there is no change in π₯. So therefore, the distance is just gonna be equal to the square root of π¦ plus five all squared. And itβs π¦ plus five because actually we could have π¦ minus negative five. So it gives us π¦ plus five.
Okay, great, so weβve now got the distance between the directrix and π₯, π¦ and the focus and π₯, π¦. So we can now look back to our relationship between our focus and directrix because actually the distance of any point on the parabola from the focus is equal to the distance from the directrix. So therefore, we know that our distances are gonna be equal. So we can actually equate them.
So we have root π₯ plus one all squared plus π¦ plus three all squared is equal to the root of π¦ plus five all squared. So therefore, if we actually square both sides of our equation, we get π₯ plus one all squared plus π¦ plus three all squared equals π¦ plus five all squared. So next, we actually expand the parentheses. So we get π₯ squared plus two π₯ plus one plus π¦ squared plus six π¦ plus nine is equal to π¦ squared plus 10π¦ plus 25. And then, we can actually subtract π¦ squared from each side, which gives us π₯ squared plus two π₯ plus six π¦ plus 10 equals 10π¦ plus 25. And then, we subtract six π¦ from each side, which gives us π₯ squared plus two π₯ plus 10 equals four π¦ plus 25.
And then, if we look back at the question, we can see that it wants us to leave it in the form π¦ equals ππ₯ squared plus ππ₯ plus π. So what weβre now gonna do is subtract 25 from each side to leave the π¦ term on its own, which gives us π₯ squared plus two π₯ minus 15 equals four π¦.
Okay, great, but have we finished yet? Well, no, if we again look back at how we want the answer left, we want the answer left in terms of π¦ and thatβs single π¦. So there is one more step that we need to do to actually reach that stage. So then, the final step is to actually divide three by four. And when we do this, weβre left with π¦ is equal to a quarter π₯ squared plus a half π₯ minus 15 over four. And this is the solution to the problem where we needed to find the equation of the parabola with a focus of negative one, negative three and directrix of π¦ equals negative five.