Answer:
[tex]\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1[/tex]
Step-by-step explanation:
Given:
Center of the ellipse is, [tex](h,k)=(-4,-3)[/tex]
Minor axis length is, [tex]2b=6[/tex]
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:
[tex]a=\sqrt{(1+4)^2+(-3+3)^2}=5[/tex]
Therefore, the value of half of major axis is, [tex]a=5[/tex]. Also,
[tex]2b=6\\b=\frac{6}{2}=3[/tex]
Now, equation of an ellipse with center [tex](h,k)[/tex] is given as:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
Plug in [tex]h=-4,k=-3,a=5,b=3[/tex] and determine the equation.
[tex]\frac{(x-(-4))^2}{5^2}+\frac{(y-(-3))^2}{3^2}=1\\\\\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1[/tex]
Therefore, the equation of the ellipse is:
[tex]\frac{(x+4)^2}{25}+\frac{(y+3)^2}{9}=1[/tex]
