conics

angeldri · #1 $Old$ 06-25-2007, 02:03 PM

Im currently in Algebra and my teacher reccommened this site, i am having problems with parabolas and ellipses and circles....aghh!!! lol

so i have sum questions

what is the directrix of the parabola in this equation: y+3 = 1/10(x+2)^2 ?

what is the center of the ellipse with this equation:
9x^2 + 16y^2 -18x +64y=71

what is the foci of this equation?
7(x-2)^2 + 3(y-2)^2=21

the vertices of this equation are: (x+3)^2- 4(y-2)^2=4

find the foci of this hyperbola: 9y^2-72y-16x^2-64x-64=0

Any help will be appreciated....thnx

Jhevon · #2 $Old$ 06-25-2007, 07:33 PM

Personally i find questions like this annoying for some reason, but you can't escape them. I'm going to try to make it as painless as possible for you.

Concerning parabolas, ellipses, hyperbolas and their vertices and foci and all that good stuff, here is what you need to know. It is one of those things where it is actually less painless to memorize and apply the formulas than trying to understand where they come from, at least as far as i'm concerned.

Parabola

The equation of a parabola with a vertical axis (upward or downward opening) can be expressed as

$y = a(x - h)^2 + k$

This is called the "Standard Form".

When in this form, the following hold:

Vertex: $(h,k)$
Focus: $\left( h,k + \frac 1{4a} \right)$
Directrix: $y = k - \frac 1{4a}$

If $a>0$ the parabola opens up
If $a<0$ the parabolas opens down

The equation of a parabola with a horizontal axis (rightward or leftward opening) can be expressed as

$x = a(y - h)^2 + k$

This is called the "Standard Form".

When in this form, the following hold:

Vertex: $(k,h)$
Focus: $\left( k + \frac 1{4a},h \right)$
Directrix: $x = k - \frac 1{4a}$

If $a>0$ the parabola opens to the right
If $a<0$ the parabola opens to the left

Alternative formulation for parabolas

The equation of a parabola with a vertical axis (upward or downward opening) can be expressed as

$(x - h)^2 = 4p(y - k)$

When in this form, the following hold:

Vertex: $(h,k)$
Focus: $(h,k + p)$
Directrix: $y = k - p$

If $p>0$ the parabola opens up
If $p<0$ the parabolas opens down

The equation of a parabola with a horizontal axis (rightward or leftward opening) can be expressed as

$(y - k)^2 = 4p(x - h)$

When in this form, the following hold:

Vertex: $(h,k)$
Focus: $(h + p,k)$
Directrix: $x = h - p$

If $p>0$ the parabola opens to the right
If $p<0$ the parabola opens to the left

Note that in the alternative formulation, $p = \frac 1{4a}$ as in the first formulation

Ellipse

The equation of an ellipse with a vertical major axis can be expressed as

$\frac {(x - h)^2}{b^2} + \frac {(y - k)^2}{a^2} = 1 \mbox { for } \boxed {a \geq b > 0}$

When in this form, the following hold:

Center: $(h,k)$
Vertices: $(h,k \pm a)$
Foci: $(h, k \pm c)$
........where $c^2 = a^2 - b^2$

The equation of an ellipse with a horizontal major axis can be expressed as

$\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1 \mbox { for } \boxed {a \geq b > 0}$

When in this form, the following hold:

Center: $(h,k)$
Vertices: $(h \pm a,k)$
Foci: $(h \pm c, k)$
........where $c^2 = a^2 - b^2$

Hyperbola

The equation of an East-West opening hyperbola can be expressed as

$\frac {(x - h)^2}{a^2} - \frac {(y - k)^2}{b^2} = 1$

When in this form, the following hold:

Center: $(h,k)$
Vertices: $(h \pm a,k)$
Asymptotes: $y = \pm \frac {b}{a} (x - h) + k$
Foci: $(h \pm c, k)$
........where $c^2 = a^2 + b^2$

The equation of an North-South opening hyperbola can be expressed as

$\frac {(y - k)^2}{a^2} - \frac {(x - h)^2}{b^2} = 1$

When in this form, the following hold:

Center: $(h,k)$
Vertices: $(h, k \pm a)$
Asymptotes: $y = \pm \frac {a}{b} (x - h) + k$
Foci: $(h, k \pm c)$
........where $c^2 = a^2 + b^2$

Circle

The equation of a circle can be expressed in the form

$(x - h)^2 + (y - k)^2 = r^2$

where the center is $(h,k)$ and the radius is $r$

Now, finally, on to your questions, so you can see how we apply the formulas above. Our objective will be to get any expression we are given into one of those forms, so we can use the rules

Quote:

Originally Posted by angeldri $View Post$

what is the directrix of the parabola in this equation: y+3 = 1/10(x+2)^2 ?

We want to get the equation given in the form $(x - h)^2 = 4p(y - k)$ so we can apply the formulas.

$y + 3 = \frac {1}{10} (x + 2)^2$

$\Rightarrow 10(y + 3) = (x + 2)^2$ .........i multiplied both sides by 10

$\Rightarrow 4 \left( \frac {10}{4} \right) (y + 3) = (x + 2)^2$ .......i rewrote the 10 in the form 4p, now we have the desired form

clearly we see that $p = \frac {10}{4}$ , $k = -3$ , and $h = -2$

Directrix: $y = k - p$

$\Rightarrow y = -3 - \frac {10}{4}$

$\Rightarrow \boxed { y = - \frac {11}{2}}$ <--------Directrix

Quote:

what is the center of the ellipse with this equation:
9x^2 + 16y^2 -18x +64y=71

We want to get the equation given in the form $\frac {(x - h)^2}{b^2} + \frac {(y - k)^2}{a^2} = 1$ or $\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1$ so we can apply the formulas

We can get this form by completing the square. I will assume you have no problem with completing the square.

$9x^2 + 16y^2 - 18x + 64y = 71$ .......let's rearrange the terms a bit

$\Rightarrow 9x^2 - 18x + 16y^2 + 64y = 71$

$\Rightarrow 9 \left( x^2 - 2x \right) + 16 \left( y^2 + 4y \right) = 71$

$\Rightarrow 9 \left[ x^2 - 2x + (-1)^2 - (-1)^2 \right] + 16 \left[ y^2 + 4y + (2)^2 - (2)^2 \right] = 71$

$\Rightarrow 9 \left[ (x - 1)^2 - 1 \right] + 16 \left[ (y + 2)^2 - 4 \right] = 71$

$\Rightarrow 9 (x - 1)^2 + 16(y + 2)^2 - 9 - 64 = 71$

$\Rightarrow 9(x - 1)^2 + 16(y + 2)^2 = 144$

$\Rightarrow \frac {(x - 1)^2}{16} + \frac {(y + 2)^2}{9} = 1$

clearly we see that $a = 4$ , $b = 3$ , $h = 1$ , and $k = -2$

So the center is $(h,k) = \boxed { (1,-2) }$

Quote:

what is the foci of this equation?
7(x-2)^2 + 3(y-2)^2=21

We want to get this equation in the form $\frac {(x - h)^2}{b^2} + \frac {(y - k)^2}{a^2} = 1$ or $\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1$

It's not hard to do here, just divide both sides by 21

$7(x - 2)^2 + 3(y - 2)^2 = 21$

$\Rightarrow \frac {(x - 2)^2}{3} + \frac {(y - 2)^2}{7} = 1$

clearly we see that $a = \sqrt {7}$ , $b = \sqrt {3}$ , $h = 2$ , $k = 2$

also, $c = \sqrt {a^2 - b^2} = \sqrt {7 - 3} = \sqrt {4} = 2$

The foci are given by: $(h, k \pm c)$

$\Rightarrow \mbox { Foci } = (2, 2 \pm 2) = \boxed {(2,4)} \mbox { and } \boxed {(2,0)}$

I'm beginning to feel that I'm spoiling you. Why not try the last two on your own. Tell me your solutions when you're done

Quote:

the vertices of this equation are: (x+3)^2- 4(y-2)^2=4

find the foci of this hyperbola: 9y^2-72y-16x^2-64x-64=0