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:Problem 1: [tex]1=\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}[/tex]Problem 2: [tex]x=\sqrt{21}\tan(\theta)[/tex][tex]y=2\sec(\theta)[/tex]Step-by-step explanation:I'm going to make a guess here.That this is two problems.Problem 1:Eliminate the parameter in [tex]x=h+b\tan(\theta)[/tex]and[tex]y=k+a\sec(theta)[/tex].Recall the following Pythagorean Identity:[tex]\tan^2(theta)+1=\sec^2(theta)[/tex]If we solve our first equation: [tex]x=h+b\tan(theta)[/tex] for the tangent piece we can substitute into this identity.  We will also need to solve for the secant piece in [tex]y=k+a\sec(\theta)[/tex].Let's do the equation for x first:[tex]x=h+b\tan(\theta)[/tex]Subtract h on both sides:[tex]x-h=b\tan(\theta)[/tex]Divide both sides by b:[tex]\frac{x-h}{b}=\tan(theta)[/tex].Let's now look at the equation for y:[tex]y=k+a\sec(\theta)[/tex]Subtract k on both sides:[tex]y-k=a\sec(\theta)[/tex]Divide boht sides by a:[tex]\frac{y-k}{a}=\sec(theta)[/tex].Now we are going to substitute this into our identity:[tex]\tan^2(theta)+1=\sec^2(theta)[/tex][tex](\frac{x-h}{b})^2+1=\(\frac{y-k}{a})^2[/tex]We are going to subtract the tangent squared piece on both sides to get 1 by itself:[tex]1=\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}[/tex]This an equation for an hyperbola with center [tex](h,k)[/tex].The vertices are at [tex](h,k\pm a)[/tex].The foci is at [tex](h,k \pm c)[/tex] where [tex]c=\sqrt{a^2+b^2}[/tex].Problem 2:We want the equation parametric equations for the conic given:Hyperbola with vertices [tex](0,\pm 2)[/tex] and foci [tex](0,\pm 5)[/tex] 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.[tex]c^2=a^2+b^2[/tex][tex]5^2=2^2+b^2[/tex][tex]25=4+b^2[/tex]Subtract 4 on both sides:[tex]21=b^2[/tex]Square root both sides:[tex]\sqrt{21}=b[/tex]Let's put into our formula for the parametric equations now:[tex]x=0+\sqrt{21}\tan(\theta)[/tex][tex]y=0+2\sec(\theta)[/tex].We don't really need that 0 in there:[tex]x=\sqrt{21}\tan(\theta)[/tex][tex]y=2\sec(\theta)[/tex]