Locus and the Parabola! (1 Viewer)

jnney

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Good evening,

The latus rectum of a parabola has endpoints (-2, 3) and (6,3). Find two possible equations for the parabola.

:)

I found the focus to be (2,3). That's all.
 

Hermes1

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Good evening,

The latus rectum of a parabola has endpoints (-2, 3) and (6,3). Find two possible equations for the parabola.

:)

I found the focus to be (2,3). That's all.
you ll have two parabolas, once concave up and one concave down
 

jnney

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How did you find 'a' and the y intercept?
 

SpiralFlex

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The distance will be 8 from point to point.

So, (Remembering the latus rectum is 4 times the focal length.)

. Our focal length.


We can see that the focus is at the midpoint of and .

The midpoint is. [Note: This is our focus.]

We know that our focal length is 2.


We also know that the latus rectum is parallel to the x axis,

So it will have the form,




Let us consider the positive case first,



So we know that the focus is 2 units from the vertex. If it is concave up, it will be two units down.

Hence, vertex is at

Our equation will be,




Consider the negative case,



Since it is concave down, we would expect the vertex to be above the focus by 2 units.

Hence the vertex is .

Therefore our equation is,

 
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Hermes1

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How did you find 'a' and the y intercept?
you know how you have the parabola in the form x^2 = 4ay

well the 4a is the length of the latus rectum.

here you can see that the latus rectum is the length of the interval from (6, 3) to (-2,3)
hence the latus rectum is 8 units.

also you know that a = focal length (distance between vertex and focus).

4a = 8
a = 2

focus (2,3) therefore one vertex is at (2,1) and another at (2, 5)

and these parabolas are in the form (x-h)^2 = 4a(y-k) where 4a is the latus rectum and (h, k) is the vertex

therefore you have (x-2)^2 = 8(y-1)
(x-2)^2 = -8(y-5)
 

Hermes1

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The distance will be 8 from point to point.

So, (Remembering the latus rectum is 4 times the focal length.)

. Our focal length.


We can see that the focus is at the midpoint of and .

The midpoint is. [Note: This is our focus.]

We know that our focal length is 2.


We also know that the latus rectum is parallel to the x axis,

So it will have the form,




Let us consider the positive case first,



So we know that the focus is 2 units from the vertex. If it is concave up, it will be two units down.

Hence, vertex is at

Our equation will be,




Consider the negative case,



Since it is concave down, we would expect the vertex to be above the focus by 2 units.

Hence the vertex is .

Therefore our equation is,

this. man latex looks so much better.
 

SpiralFlex

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It also takes much longer. I might stop helping now.
 

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