To find the rectangular form of the given parametric equations [tex]\( x = 4 \cos(t) - 3 \)[/tex] and [tex]\( y = 5 \sin(t) + 2 \)[/tex], we follow these steps:
1. Isolate the trigonometric functions [tex]\(\cos(t)\)[/tex] and [tex]\(\sin(t)\)[/tex]:
[tex]\[
\cos(t) = \frac{x + 3}{4}
\][/tex]
[tex]\[
\sin(t) = \frac{y - 2}{5}
\][/tex]
2. Apply the Pythagorean identity which states:
[tex]\[
\sin^2(t) + \cos^2(t) = 1
\][/tex]
3. Substitute [tex]\(\cos(t)\)[/tex] and [tex]\(\sin(t)\)[/tex] with their expressions from step 1:
[tex]\[
\left( \frac{y - 2}{5} \right)^2 + \left( \frac{x + 3}{4} \right)^2 = 1
\][/tex]
4. Rewrite the equation in standard form by squaring the terms in the equation:
[tex]\[
\left( \frac{y - 2}{5} \right)^2 = \frac{(y - 2)^2}{25}
\][/tex]
[tex]\[
\left( \frac{x + 3}{4} \right)^2 = \frac{(x + 3)^2}{16}
\][/tex]
So, the equation becomes:
[tex]\[
\frac{(y - 2)^2}{25} + \frac{(x + 3)^2}{16} = 1
\][/tex]
5. Reorder the terms to match the standard form of an ellipse equation [tex]\(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\)[/tex]. From the equation we derived:
[tex]\[
\frac{(x + 3)^2}{16} + \frac{(y - 2)^2}{25} = 1
\][/tex]
From these steps, the correct rectangular form of the given parametric equations is:
[tex]\[
\boxed{\frac{(x+3)^2}{16}+\frac{(y-2)^2}{25}=1}
\][/tex]
