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,

\(\displaystyle\frac{{4}^{2}}{{a}^{2}}+\frac{{3}^{2}}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}={1}-\frac{9}{{b}^{2}}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Step 2

Equation of the ellipse through the point (6, 2).

So, \(\displaystyle{x}={6},{y}={2}\) satisfies the equation of an ellipse.

\(\displaystyle\Rightarrow\frac{{6}^{2}}{{a}^{2}}+\frac{{2}^{2}}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}+\frac{4}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}={1}-\frac{4}{{b}^{2}}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}\)

Step 3

Compare equations (1) and (2), we get

\(\displaystyle\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Multiply both sides by 4,

\(\displaystyle\frac{1}{{9}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{4}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Distribute both sides,

\(\displaystyle\frac{1}{{9}}-\frac{4}{{{9}{b}^{2}}}=\frac{1}{{4}}-\frac{9}{{{4}{b}^{2}}}\)

Combine like terms

\(\displaystyle\frac{1}{{9}}-\frac{1}{{4}}=\frac{4}{{{9}{b}^{2}}}-\frac{9}{{{4}{b}^{2}}}\)

Make the same denominator,

\(\displaystyle\frac{4}{{36}}-\frac{9}{{36}}=\frac{16}{{{36}{b}^{2}}}-\frac{81}{{{36}{b}^{2}}}\)

\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)

\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)

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 \(\displaystyle\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)},\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{13}}\right)}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left(\frac{4}{{13}}\right)}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{4}}{\left(\frac{1}{{13}}\right)}\)

Taking reciprocal from both sides

\(\displaystyle\Rightarrow{a}^{2}={52}\)

The equation of ellipse through the points (4, 3) and (6, 2) is \(\displaystyle\frac{{x}^{2}}{{52}}+\frac{{y}^{2}}{{13}}={1}\).

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,

\(\displaystyle\frac{{4}^{2}}{{a}^{2}}+\frac{{3}^{2}}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{16}{{a}^{2}}={1}-\frac{9}{{b}^{2}}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Step 2

Equation of the ellipse through the point (6, 2).

So, \(\displaystyle{x}={6},{y}={2}\) satisfies the equation of an ellipse.

\(\displaystyle\Rightarrow\frac{{6}^{2}}{{a}^{2}}+\frac{{2}^{2}}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}+\frac{4}{{b}^{2}}={1}\)

\(\displaystyle\Rightarrow\frac{36}{{a}^{2}}={1}-\frac{4}{{b}^{2}}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}\)

Step 3

Compare equations (1) and (2), we get

\(\displaystyle\frac{1}{{36}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Multiply both sides by 4,

\(\displaystyle\frac{1}{{9}}{\left({1}-\frac{4}{{b}^{2}}\right)}=\frac{1}{{4}}{\left({1}-\frac{9}{{b}^{2}}\right)}\)

Distribute both sides,

\(\displaystyle\frac{1}{{9}}-\frac{4}{{{9}{b}^{2}}}=\frac{1}{{4}}-\frac{9}{{{4}{b}^{2}}}\)

Combine like terms

\(\displaystyle\frac{1}{{9}}-\frac{1}{{4}}=\frac{4}{{{9}{b}^{2}}}-\frac{9}{{{4}{b}^{2}}}\)

Make the same denominator,

\(\displaystyle\frac{4}{{36}}-\frac{9}{{36}}=\frac{16}{{{36}{b}^{2}}}-\frac{81}{{{36}{b}^{2}}}\)

\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)

\(\displaystyle\Rightarrow-\frac{5}{{36}}=-\frac{65}{{{36}{b}^{2}}}\)

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 \(\displaystyle\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{b}^{2}}\right)},\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left({1}-\frac{9}{{13}}\right)}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{16}}{\left(\frac{4}{{13}}\right)}\)

\(\displaystyle\Rightarrow\frac{1}{{a}^{2}}=\frac{1}{{4}}{\left(\frac{1}{{13}}\right)}\)

Taking reciprocal from both sides

\(\displaystyle\Rightarrow{a}^{2}={52}\)

The equation of ellipse through the points (4, 3) and (6, 2) is \(\displaystyle\frac{{x}^{2}}{{52}}+\frac{{y}^{2}}{{13}}={1}\).