Answer
$(a)$ The closest point is $(-2,0)$, highlighted as the left red dot on the image below. And the farthest point is $(8,0)$, the right red dot on the image.
$(b)$ $x$-coordinates are $-1.33$ and $7.33$
Distances are $2.4 Mm$ and $7.6Mm$
Work Step by Step
$(a)$ As we can see from the image, the closest point is $(-2,0)$, highlighted as the left red dot on the image below. And the farthest point is $(8,0)$, the right red dot on the image.
$(b)$ We know the equation of satellite's orbit:
$\frac{(x-3)^2}{25}+\frac{y^2}{16}=1$
We need to find $x$-coordinates of these points, where $y=2$. We will simply input $y$=value and find $x$.
$\frac{(x-3)^2}{25}+\frac{2^2}{16}=1$
$\frac{(x-3)^2}{25}+\frac{1}{4}=1$
$\frac{(x-3)^2}{25}=\frac{3}{4}$
$4(x^2-6x+9)=75$
$4x^2-24x-39=0$
$x^2-6x-9.75=0$
$D=b^2-4ac=36+39=75$
$x_1=\frac{-b-\sqrt{D}}{2a}=\frac{6-5\sqrt{3}}{2}\approx -1.33$
$x_2=\frac{-b+\sqrt{D}}{2a}=\frac{6+5\sqrt{3}}{2} \approx 7.33$
We have points $A(-1.33, 2)$ and $B(7.33, 2)$
We have to find distance $AO$ and $BO$
$AO=\sqrt{(0+1.33)^2+(0-2)^2}=\sqrt{1.77+4} \approx 2.4 Mm$
$BO=\sqrt{(0-7.33)^2+(0-2)^2}=\sqrt{53.73+4} \approx 7.6 Mm$
