Answer
a) $\dfrac{x^2}{2304}+\dfrac{y^2}{529}=1$
b) $(-5\sqrt {71},0)\approx (-42.1,0)$
$(5\sqrt {71},0)\approx (42.1,0)$
Work Step by Step
a) We are given:
$a=48$
$b=23$
Determine the equation of the ellipse:
$\dfrac{x^2}{48^2}+\dfrac{y^2}{23^2}=1$
$\dfrac{x^2}{2304}+\dfrac{y^2}{529}=1$
b) Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
$h=0$
$k=0$
$a=48$
$b=23$
Determine the foci:
$c^2=a^2-b^2$
$c^2=48^2-23^2$
$c^2=1775$
$c=\pm\sqrt {1775}=\pm 5\sqrt{71}$
The foci are:
$F_1(h-c,k)=(-5\sqrt {71},0)\approx (-42.1,0)$
$F_2(h+c,k)=(5\sqrt {71},0)\approx (42.1,0)$
Draw the equation $y=-\sqrt{4-4x^2}$, which is the half of ellipse with negative $y$.
