Hello there. To solve this question, we'll have to find the rectangular equation given the parametric equations.
The parametric equations are:
[tex]\begin{gathered} x=-7+2\cos \theta \\ y=4+5\sin \theta \end{gathered}[/tex]Let's start addind 7 on both sides of the first equation and subtracting 4 on both sides of the second, such that
[tex]\begin{gathered} x+7=2\cos \theta \\ y-4=5\sin \theta \end{gathered}[/tex]Divide both sides of the first equation by a factor of 2 and the second by a factor of 5
[tex]\begin{gathered} \frac{x+7}{2}=\cos \theta \\ \\ \frac{y-4}{5}=\sin \theta \end{gathered}[/tex]Now, we simply apply the fundamental trigonometric identity:
[tex]\cos ^2\theta+\sin ^2\theta=1[/tex]such that we have
[tex]\left(\frac{x+7}{2}\right)^2+\left(\frac{y-4}{5}\right)^2=1[/tex]Square the terms
[tex]\frac{(x+7)^2}{4}+\frac{(y-4)^2}{25}=1[/tex]This is the rectangular equation we were looking for. This is, in fact, an ellipse wih center at (-7, 4) and semi-major and semi-minor axes equal to 5 and 4, respectively.
The other way we could have solved this question is by knowing when he have an ellipse with center at (h, k) and semi-major and semi-minor axes respectively equal to a and b, the parametric equations are given by:
[tex]\begin{gathered} x=h+a\cos \theta \\ y=k+b\sin \theta \end{gathered}[/tex]If the axes changes places, that is, when the ellipse semi-major axis is parallel to the y-axis, then
[tex]\begin{gathered} x=h+b\cos \theta \\ y=k+a\sin \theta \end{gathered}[/tex]And the formulas for the ellipses are the same as before
[tex]\begin{gathered} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2_{}}{b^2}=1 \\ \text{ or} \\ \frac{(x-h)^2}{b^2^{}}+\frac{(y-k)^2_{}}{a^2}=1 \end{gathered}[/tex]