Question

Eliminate the parameter and obtain the standard form of the rectangular equation. Hyperbola: x = h + b tan(θ), y = k + a sec(θ) Use your result to find a set of parametric equations for the line or conic. (When 0 ≤ θ ≤ 2π. Set your center at the origin. Enter your answers as a comma-separated list of equations.) Hyperbola: vertices: (0, ±2); foci: (0, ± 5 )

Answer 1

Answer:

Problem 1:

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

Problem 2:

$x=\sqrt{21}\tan(\theta)$

$y=2\sec(\theta)$

Step-by-step explanation:

I'm going to make a guess here.

That this is two problems.

Problem 1:

Eliminate the parameter in

$x=h+b\tan(\theta)$

and

$y=k+a\sec(theta)$.

Recall the following Pythagorean Identity:

$\tan^2(theta)+1=\sec^2(theta)$

If we solve our first equation: $x=h+b\tan(theta)$ for the tangent piece we can substitute into this identity.  We will also need to solve for the secant piece in $y=k+a\sec(\theta)$.

Let's do the equation for x first:

$x=h+b\tan(\theta)$

Subtract h on both sides:

$x-h=b\tan(\theta)$

Divide both sides by b:

$\frac{x-h}{b}=\tan(theta)$.

Let's now look at the equation for y:

$y=k+a\sec(\theta)$

Subtract k on both sides:

$y-k=a\sec(\theta)$

Divide boht sides by a:

$\frac{y-k}{a}=\sec(theta)$.

Now we are going to substitute this into our identity:

$\tan^2(theta)+1=\sec^2(theta)$

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

We are going to subtract the tangent squared piece on both sides to get 1 by itself:

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

This an equation for an hyperbola with center $(h,k)$.

The vertices are at $(h,k\pm a)$.

The foci is at $(h,k \pm c)$ where $c=\sqrt{a^2+b^2}$.

Problem 2:

We want the equation parametric equations for the conic given:

Hyperbola with vertices $(0,\pm 2)$ and foci $(0,\pm 5)$ and center at (0,0).

We are using the information from the first sentence.

We have (h,k)=(0,0).

We have a=2 and c=5 so we need to find b.

$c^2=a^2+b^2$

$5^2=2^2+b^2$

$25=4+b^2$

Subtract 4 on both sides:

$21=b^2$

Square root both sides:

$\sqrt{21}=b$

Let's put into our formula for the parametric equations now:

$x=0+\sqrt{21}\tan(\theta)$

$y=0+2\sec(\theta)$.

We don't really need that 0 in there:

$x=\sqrt{21}\tan(\theta)$

$y=2\sec(\theta)$