Answer
$\dfrac{(x-3)^2}{25}+\dfrac{(y+2)^2}{4}=1$
$(3-\sqrt{21},-2)$
$(3+\sqrt{21},-2)$
Work Step by Step
We are given the equation:
$4x^2+25y^2-24x+100y+36=0$
Form two squares:
$(4x^2-24x+36)+(25y^2+100y+100)-36-100+36=0$
$4(x^2-6x+9)+25(y^2+4y+4)-100=0$
$4(x-3)^2+25(y+2)^2=100$
Divide both sides by 100:
$\dfrac{4(x-3)^2}{100}+\dfrac{25(y+2)^2}{100}=1$
Simplify:
$\dfrac{(x-3)^2}{25}+\dfrac{(y+2)^2}{4}=1$
$\dfrac{(x-3)^2}{5^2}+\dfrac{(y+2)^2}{2^2}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
$h=3$
$k=-2$
$a=5$
$b=2$
Determine the foci:
$c^2=a^2-b^2$
$c^2=5^2-2^2$
$c^2=21$
$c=\pm\sqrt {21}$
The foci are:
$F_1(h-c,k)=(3-\sqrt{21},-2)$
$F_2(h+c,k)=(3+\sqrt{21},-2)$
