1. parabola ques practice

y = 4x^2 + 8x + 5

find the focal length, the focus and the directrix.

any help? much appreciated

2. Originally Posted by jvignacio
y = 4x^2 + 8x + 5

find the focal length, the focus and the directrix.

any help? much appreciated
see post #2 here

note, the focal length is the distance between the focus and the vertex of the parabola, the post tells you how to find both the vertex and focus, so just use the distance formula from there

3. Originally Posted by jvignacio
y = 4x^2 + 8x + 5

find the focal length, the focus and the directrix.

any help? much appreciated
Hey jv,

There is more than one way to approach this, so here is one way. Remember that a parabola is a curve consisting of all points in the coordinate plane that are the same distance from a given point (the focus) and a given line (the directrex). The focal length is the distance from the vertex to the focus.

Examples with solutions can be found here, as well as a number of other places on the net.

$y = 4x^2 + 8x + 5$

First, put the equation in this form: $y=a(x-h)^2+k$

You'll have to know how to complete the square. Do you?

I'll get you started:

$y=4(x^2+2x+ ??)+5-4(??)$

Once you have it in the above form, apply the following information to find what you need:

Vertex $(h, k)$

Focus $\left(h, k+\frac{1}{4a}\right)$

Directrex $y=k-\frac{1}{4a}$

4. Originally Posted by masters
Hey jv,

There is more than one way to approach this, so here is one way. Remember that a parabola is a curve consisting of all points in the coordinate plane that are the same distance from a given point (the focus) and a given line (the directrex). The focal length is the distance from the vertex to the focus.

Examples with solutions can be found here, as well as a number of other places on the net.

$y = 4x^2 + 8x + 5$

First, put the equation in this form: $y=a(x-h)^2+k$

You'll have to know how to complete the square. Do you?

I'll get you started:

$y=4(x^2+2x+ ??)+5-4(??)$

Once you have it in the above form, apply the following information to find what you need:

Vertex $(h, k)$

Focus $\left(h, k+\frac{1}{4a}\right)$

Directrex $y=k-\frac{1}{4a}$
much appreciated for taking your time to explain it. thank u.. ill giv it a shot and post it back to check

5. Originally Posted by masters
Hey jv,

There is more than one way to approach this, so here is one way. Remember that a parabola is a curve consisting of all points in the coordinate plane that are the same distance from a given point (the focus) and a given line (the directrex). The focal length is the distance from the vertex to the focus.

Examples with solutions can be found here, as well as a number of other places on the net.

$y = 4x^2 + 8x + 5$

First, put the equation in this form: $y=a(x-h)^2+k$

You'll have to know how to complete the square. Do you?

I'll get you started:

$y=4(x^2+2x+ ??)+5-4(??)$

Once you have it in the above form, apply the following information to find what you need:

Vertex $(h, k)$

Focus $\left(h, k+\frac{1}{4a}\right)$

Directrex $y=k-\frac{1}{4a}$
my final formula was.. 4(x-1)^2 + 1

when i substituted the values.. i got the vertex = (1,1) directrex = 15/16 and focus (1, 17/16) i think i did something wrong??

6. Originally Posted by jvignacio
my final formula was.. 4(x-1)^2 + 1

when i substituted the values.. i got the vertex = (1,1) directrex = 15/16 and focus (1, 17/16) i think i did something wrong??
Almost. The vertex is (-1, 1).

That makes the directrex equation $y=\frac{15}{16}$.

The focus is $\left(-1, \frac{17}{16}\right)$

7. Originally Posted by jvignacio
my final formula was.. 4(x-1)^2 + 1

when i substituted the values.. i got the vertex = (1,1) directrex = 15/16 and focus (1, 17/16) i think i did something wrong??
$y=4x^2+8x+5$

$y=4(x^2+2x+1)+5-4$

$y=4(x+1)^2+1$

Vertex (-1, 1)

8. Originally Posted by masters
$y=4x^2+8x+5$

$y=4(x^2+2x+1)+5-4$

$y=4(x+1)^2+1$

Vertex (-1, 1)
thank u for clearing that up.
much appreciated

9. Originally Posted by jvignacio
thank u for clearing that up.
much appreciated
You're welcome!