# alg 2 conics

• Apr 8th 2007, 09:01 PM
mattballer082
alg 2 conics
if possible help me step by step. i need to understand this.

Find the equation of the hyperbola described:

(a) foci (0, +-3) asymptotes y = +- 3/4

(b) center (-1, 3) vertices (-4,3) and (2,3) foci (-6,3) and (4,3)..

very confused. anyone?

also how is this problem done with a circle and a parabola?

the circle X (squared) + Y (squared) = 4 and parabola 5y = -3x (squared). intersect at the points:????
how do i find the points?
• Apr 9th 2007, 11:24 AM
topsquark
Quote:

Originally Posted by mattballer082
Find the equation of the hyperbola described:

(a) foci (0, +-3) asymptotes y = +- 3/4

The asymptotes indicate that this is not a hyperbola. Is there a typo here?

Quote:

Originally Posted by mattballer082
also how is this problem done with a circle and a parabola?

the circle X (squared) + Y (squared) = 4 and parabola 5y = -3x (squared). intersect at the points:????
how do i find the points?

x^2 + y^2 = 4
5y = -3x^2

Typically we would solve the bottom equation for x or y and plug it into the top equation. But the bottom equation is practically solved anyway, so no trouble there:
5y = -3x^2 --> y = -(3/5)x^2

Insert this into the circle equation:
x^2 + y^2 = 4

x^2 + [-(3/5)x^2]^2 = 4

Now solve for x:
x^2 + 9/25 * x^4 = 4

9/25 * x^4 + x^2 - 4 = 0 <-- Multiply both sides by 25

9x^4 + 25x^2 - 100 = 0

This is what is called a "biquadratic equation." Note that only even powers of x show up. I'm going to define
z = x^2

Then
9z^2 + 25z - 100 = 0

z = [-25 (+/-) sqrt{25^2 - 4*9*(-100)}]/(2*9)

z = [-25 (+/-) sqrt{625 + 3600}]/18

z = [-25 (+/-) sqrt{4225}]/18

z = [-25 (+/-) sqrt{4225}]/18

z = [-25 (+/-) 65]/18

So z = 40/18 = 20/9 or z = -90/18 = -5.

Thus
z = x^2 implies that:
x = (+/-)sqrt{20/9} or x = (+/-)sqrt{-5}

So
x = (+/-)(2/3)*sqrt{5}
are the only two x values. (I'm ignoring the z = -5 solution because we can't take the square root of a negative number.)

So what are the corresponding y values? Well:
y = -(3/5)x^2

y = -(3/5)*[(+/-)(2/3)*sqrt{5}] = -(3/5)(20/9) = -4/3 (for both the + and - solutions, of course.)

So there are two intersection points:
(x, y) = (2/3*sqrt{5}, -4/3)
and
(x, y) = (-2/3*sqrt{5}, -4/3)

I have attached a graph of both curves below.

-Dan