Solve for the equation in standard form of the following conic sections and graph the...

Step 1
An equation of the ellipse is of the form,
${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$
Equation of the ellipse through the point (4, 3).
So, ${\displaystyle x=4,y=3}$ satisfies the equation of an ellipse.
Implies that,
${\displaystyle \frac{4^2}{a^2}+\frac{3^2}{b^2}=1}$
${\displaystyle \Rightarrow \frac{16}{a^2}+\frac{9}{b^2}=1}$
${\displaystyle \Rightarrow \frac{16}{a^2}=1-\frac{9}{b^2}}$
${\displaystyle \Rightarrow \frac{1}{a^2}=\frac{1}{16}(1-\frac{9}{b^2})}$
Step 2
Equation of the ellipse through the point (6, 2).
So, ${\displaystyle x=6,y=2}$ satisfies the equation of an ellipse.
${\displaystyle \Rightarrow \frac{6^2}{a^2}+\frac{2^2}{b^2}=1}$
${\displaystyle \Rightarrow \frac{36}{a^2}+\frac{4}{b^2}=1}$
${\displaystyle \Rightarrow \frac{36}{a^2}=1-\frac{4}{b^2}}$
${\displaystyle \Rightarrow \frac{1}{a^2}=\frac{1}{36}(1-\frac{4}{b^2})}$
Step 3
Compare equations (1) and (2), we get
${\displaystyle \frac{1}{36}(1-\frac{4}{b^2})=\frac{1}{16}(1-\frac{9}{b^2})}$
Multiply both sides by 4,
${\displaystyle \frac{1}{9}(1-\frac{4}{b^2})=\frac{1}{4}(1-\frac{9}{b^2})}$
Distribute both sides,
${\displaystyle \frac{1}{9}-\frac{4}{9b^2}=\frac{1}{4}-\frac{9}{4b^2}}$
Combine like terms
${\displaystyle \frac{1}{9}-\frac{1}{4}=\frac{4}{9b^2}-\frac{9}{4b^2}}$
Make the same denominator,
${\displaystyle \frac{4}{36}-\frac{9}{36}=\frac{16}{36b^2}-\frac{81}{36b^2}}$
${\displaystyle \Rightarrow -\frac{5}{36}=-\frac{65}{36b^2}}$
${\displaystyle \Rightarrow -\frac{5}{36}=-\frac{65}{36b^2}}$
Multiply both sides by 36,
${\displaystyle \Rightarrow -\frac{5}{1}=-\frac{65}{b^2}}$
Divide both sides by -65,
${\displaystyle \Rightarrow \frac{5}{65}=\frac{1}{b^2}}$
By simplifying it,
${\displaystyle \Rightarrow \frac{1}{13}=\frac{1}{b^2}}$
Taking reciprocal from both sides,
${\displaystyle \Rightarrow b^2=13}$
Step 4
Substitute ${\displaystyle b^2=13}$
in the equation ${\displaystyle \frac{1}{a^2}=\frac{1}{16}(1-\frac{9}{b^2}),}$