If a parabola has the focus at (2,4)\left(2,4\right)(2,4) and a directrix line x=−4x=-4x=−4 , find the standard equation for the parabola.

12 ( x + 1 ) = ( y − 4 ) 2 12\left(x+1\right)=\left(y-4\right)^2 12 ( x + 1 ) = ( y − 4 ) 2

If a parabola has the focus at ( 2 , 4 ) \left(2,4\right) ( 2 , 4 ) and a directrix line x = − 4 x=-4 x = − 4 , find the standard equation for the parabola.

Hey, everyone. Let's see if we can do this problem. So in this problem, we are told if a parabola has the focus point at 24, and a directrix line at x equals negative 4, find the standard equation for the parabola. Now, what I'm going to do to solve this is I'm first going to draw out a graph, and we don't actually have to graph this parabola, but this will kind of give us some reference as to what we're dealing with here. So we'll have a standard x and y graph, What I'm going to do is draw where about the focus and directrix would be. Well, you can see the focus is at the point 24, so that means horizontally we're 1, 2 units to the right, and vertically we're up 4 units. So 1, 2, 3, 4, this would be our focus. So if this is the focus point right here, we're told the directrix line is at x equals negative 4. So that means we need to go back 1, 2, 3, 4 units to find our directrix line, which would look something like that. So this is what we would have and this would be our directrix line, and we would say that this is x equals negative 4, and we know that our focus point is at 24, and so our parabola is going to be somewhere in the middle here. And if I look, this is actually the the center point of our parabola, because notice it takes 1, 2, 3 units to get there, and takes 1, 2, 3 units to get there. So this is right in the middle of our directrix and focus. Now the question also is gonna become what direction the parabola opens. And I can tell it's gonna open to the right, because that's where our focus is. And our parabola and focus always open or go in the same direction. So since our parabola opens this way, that's why the focus would be there, or we saw the focus was already there, so our parabola must open that way. Now the standard equation for a sideways parabola like this is 4p times x minus h is equal to y minus k squared, and what I can do is figure out that our vertex here well our vertex is at a horizontal position of negative one, and a vertical position of 4. And so because this is our vertex that tells us what the 'h' and 'k' are going to be. Our 'h' is going to be negative one, and our 'k' is going to be 4. Now, the last thing I need to actually solve this, so we only have x's and y's left in the equation is to figure out what this p value is, and recall that p is the or the absolute value of p is the distance to both the to both the focus point as well as the directrix, and since I can see that we have to go 1, 2, 3 units to get to the focus point, and 1, 2, 3 units to the directrix the absolute value of p is 3, and the p value is also just gonna be positive 3, because we can know that this opens to the right, so we don't have to worry about any negative numbers. So what that means is that our p value is going to be 3. So plugging everything into this equation, we're gonna have 4 times 3 times x minus negative one, and then this is going to equal y minus 4 squared, and really I should have bigger brackets here. So this is what we have, and 4 times 3 that's 12, and then we can cancel the negatives here to give us x plus 1, and then this is equal to y minus 4 squared. So this is what the equation would look like given the focus and directrix. So notice how we were able to take the focus and directrix already given to us, and construct the standard equation for the parabola that we are dealing with and this is how we can solve these types of problems when we're not at the origin and we have a sideways parabola. So hope you found this helpful. Thanks for watching, and let me know if you have any questions.

If a parabola has the focus at ( 2 , 4 ) \left(2,4\right) ( 2 , 4 ) and a directrix line x = − 4 x=-4 x = − 4 , find the standard equation for the parabola.

12 ( x + 1 ) = ( y − 4 ) 2 12\left(x+1\right)=\left(y-4\right)^2 12 ( x + 1 ) = ( y − 4 ) 2

− ( x + 1 ) = ( y − 4 ) 2 -\left(x+1\right)=\left(y-4\right)^2 − ( x + 1 ) = ( y − 4 ) 2

4 ( x − 1 ) = ( y + 4 ) 2 4\left(x-1\right)=\left(y+4\right)^2 4 ( x − 1 ) = ( y + 4 ) 2

8. Conic Sections

Parabolas

College Algebra

8. Conic Sections

Parabolas

Multiple Choice

If a parabola has the focus at $\left(2,4\right)$ and a directrix line $x=-4$ , find the standard equation for the parabola.

$12\left(x+1\right)=\left(y-4\right)^2$

$-\left(x+1\right)=\left(y-4\right)^2$

$12x=y^2$

$4\left(x-1\right)=\left(y+4\right)^2$

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If a parabola has the focus at ﻿(2,4)\left(2,4\right)(2,4)﻿ and a directrix line ﻿x=−4x=-4x=−4﻿ , find the standard equation for the parabola.

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If a parabola has the focus at $\left(2,4\right)$ and a directrix line $x=-4$ , find the standard equation for the parabola.