Need Reassurance on hyperbolas, ellipses, etc

[Jaskaran]

Hello, I have a few, but miniscule questions that I need reassurance about.

1) How do I find the asymptote of a hyperbola?

The equation is \(\displaystyle (X^2 / 169) - (Y^2 / 64) = 1\)

I get \(\displaystyle y=+- 8/13x\) It's y=+-b/ax right?

2) I know that to find the area of an ellipse, it's A*B*Pi, so in an equation such as:

\(\displaystyle 5x^2+ 12y^2 = 60\)

I would divide everything by 60 and multiply the square root of 12 by the square root of 5, right?

3) Seismograph shows the epicenter of an earthquake to be about 40 miles from station 1. (station 1 is at 0,0 on the coordinate grid). Another station, which is 60 miles west and 40 miles north of station 1 find that the quake is 60 miles away. Find the possible locations of the center.

Okay, for that one, in coordinate grids, or for the equation of a circle, west is on the negative side and is written as positive for the x-value right?

For example, this is how I set-it up.

\(\displaystyle x^2 + y^2 = 1600\) For station 1.

\(\displaystyle (x+60)^2 + (y-40)^2 = 3600\)

So I FOIL and solve the system right?

My work:

\(\displaystyle x^2 + y^2 = 1600\)

\(\displaystyle X^2 + 120x + 3600 + y^2 -80x + 1600 = 3600\)

Where do I go from here?

4) Last but not least: An oriental rug has area of 238ft^2 and a perimeter of 62 ft. Find the dimensions.

So I set up:

Lw = 238
2L + 2W = 62

Where to from there?
Solve for w or L?

5) Also, and lastly: In the equation:

\(\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0\)

IF B=0, what else must be true in order for the equation to represent a parabola?

I know that if \(\displaystyle B^2 - 4AC=0\) for it to be a parabola. So what would I put in order for the answer? This is confusing, please help.

 

[soroban]

Hello, Jaskaran!

You're doing fine . . . and thank you for showing your work!

Hello, I have a few, but miniscule questions that I need reassurance about.

1) How do I find the asymptote of a hyperbola?

\(\displaystyle \;\;\)The equation is \(\displaystyle \frac{x^2}{169}\,-\,\frac{y^2}{64}\;=\;1\)

I get \(\displaystyle y\:=\:\pm\frac{8}{13}x\;\;\) It's\(\displaystyle y\:=\:\pm\frac{b}{a}x\), right? \(\displaystyle \;\) . . . Right!

2) I know that to find the area of an ellipse, it's \(\displaystyle \pi ab\),
so in an equation such as: \(\displaystyle 5x^2\,+\,12y^2\:=\:60\)

I would divide everything by 60: \(\displaystyle \:\frac{x^2}{12}\,+\,\frac{y^2}{5}\:=\:1\)
and multiply \(\displaystyle \sqrt{12}\) by \(\displaystyle \sqrt{5}\), right? . . . Right!

Correct!
\(\displaystyle \;\;A\:=\:\pi(\sqrt{12})(\sqrt{5})\:=\:\pi\sqrt{60}\:=\:2\sqrt{15}\,\pi\)

3) Seismograph shows the epicenter of an earthquake to be about 40 miles from station A at (0,0).
Station B, 60 miles west and 40 miles north of A. finds that the quake is 60 miles away.
Find the possible locations of the center.

For example, this is how I set-it up.

Station A: \(\displaystyle \,x^2\,+\,y^2\:=\:1600\)

Station B: \(\displaystyle \,(x+60)^2\,+\,(y-40)^2\:=\:3600\)

So I FOIL and solve the system right?

My work:

\(\displaystyle x^2\,+\,y^2\:=\:1600\)
\(\displaystyle x^2\,+\,120x\,+\,3600\,+\,y^2\.-\,80x\,+\,1600\:=\:3600\;\) . . . Yes!

Where do I go from here?

We have a system of quadratic equations to solve . . . always messy.

We have: \(\displaystyle \,\begin{array}{cc}[1]\;x^2\,+\,y^2\,+\,120x\,-\,80y\;=\;-1600 \\ [2]\;x^2\,+\,y^2\;=\;1600\end{array}\)

Subtract: \(\displaystyle \,120x\,-\,8y\:=\:-3200\;\;\Rightarrow\;\;y\:=\:15x\,+\,400\)

Substitute into [2]: \(\displaystyle \,x^2\,+\,(15x\,+\,400)^2\;=\;1600\)

\(\displaystyle \;\;\)and we have a quadratic equation to solve: \(\displaystyle \,226x^2\,+\,12,000x\,+\,158,400\;=\;0\)

Like I said: messy . . .

 

[soroban]

Hello again, Jaskaran!

4) An oriental rug has area of 238 ft² and a perimeter of 62 ft.
Find the dimensions.

So I set up:

LW = 238
2L + 2W = 62 \(\displaystyle \;\) . . . Good!

Where to from there?
Solve for W or L? \(\displaystyle \;\) . . . Either one

We have: \(\displaystyle \;\begin{array}{cc}[1]\;LW\:=\:238 \\ [2]\;L\,+\,W\:=\:31\end{array}\)

From [1], we have: \(\displaystyle \,W\,=\,\frac{238}{L}\)

Substitute into [2]: \(\displaystyle \,L\,+\,\frac{238}{L}\:=\:31\)

\(\displaystyle \;\;\)which simplifies to the quadratic: \(\displaystyle \,L^2\,-\,31L\,+\,238\:=\:0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \,(L\,-\,14)(L\,-\,17)\:=\:0\)

\(\displaystyle \;\;\) and has roots: \(\displaystyle \,L\:=\:14,\:17\) . . . then: \(\displaystyle \,W\:=\:17,\;14\)

Assuming that the length is greater than the width,
\(\displaystyle \;\;\)the solution is: \(\displaystyle \,L\,=\,17,\;W\,=\,14\)

5) In the equation: \(\displaystyle \,Ax^2\,+\,Bxy\,+\,Cy^2\,+\,Dx\,+\,Ey\,+\,F\;=\;0\)

IF \(\displaystyle B = 0\), what else must be true in order for the equation to represent a parabola?

I know that \(\displaystyle B^2\,-\,4AC\:=\:0\) for it to be a parabola. \(\displaystyle \;\) . . . Yes!
So what would I put in order for the answer?

We know that \(\displaystyle B\,=\,0\)
\(\displaystyle \;\;\)so the equation becomes: \(\displaystyle \,Ax^2\,+\,Cy^2\,+\,Dx\,+\,Ey\,+\,F\:=\:0\)

The discriminant must be zero: \(\displaystyle \,0^2\,-\,4AC\:=\:0\;\;\Rightarrow\;\;AC\:=\:0\)

That is, either \(\displaystyle A\,=\,0\) or \(\displaystyle C\,=\,0\) . . . but not both.


This makes sense . . .

With a parabola, one of the variables is squared, the other is linear.
\(\displaystyle \;\;\)So either \(\displaystyle A\) or \(\displaystyle C\) must be zero.

Both cannot be zero . . . we'd have: \(\displaystyle \,Dx\,+\,Ey\,+\,F\:=\:0\) . . . a straight line.

 

[Jaskaran]

Thanks Soroban!

 

[Jaskaran]

Ooops. Also, what if C=0 , what else must be true in order for the equation to be a parabola? :lol:

B must equal 0?

Thanks.