Question

Will give brainliest plz help

Answer 1

Answer:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Step-by-step explanation:

Given:

Center of the ellipse is, $(h,k)=(-4,-3)$

Minor axis length is, $2b=6$

A vertex of the ellipse is at (1, -3)

Now, distance between the center and the vertex is half of the length of the major axis.

Using distance formula for (-4, -3) and (1, -3), we get:

$a=\sqrt{(1+4)^2+(-3+3)^2}=5$

Therefore, the value of half of major axis is, $a=5$. Also,

$2b=6\\b=\frac{6}{2}=3$

Now, equation of an ellipse with center $(h,k)$ is given as:

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

Plug in $h=-4,k=-3,a=5,b=3$ and determine the equation.

$\frac{(x-(-4))^2}{5^2}+\frac{(y-(-3))^2}{3^2}=1\\\\\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$

Therefore, the equation of the ellipse is:

$\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1$