Given:

Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Since the major axis is on the x-axis, the equation of the ellipse will be the form

\( \dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1,\)…. (1) [Where ‘a’ is the semi-major axis.]

The ellipse passes through points (4, 3) and (6, 2).

So by putting the values x = 4 and y = 3 in equation (1), we get,

\(\dfrac{ 16}{a^2} + \dfrac{9}{b^2} = 1\) …. (2)

Putting, x = 6 and y = 2 in equation (1), we get,

\(\dfrac{ 36}{a^2} + \dfrac{4}{b^2 }= 1\) …. (3)

From equation (2)

\(\dfrac{ 16}{a^2} = 1 – \dfrac{9}{b^2}\)

\( \dfrac{ 1}{a^2} = \dfrac{1}{16}(1 – \dfrac{9}{b^2})\) …. (4)

Substituting the value of \( \dfrac{ 1}{a^2} \) in equation (3) we get,

b^{2} = 13

Now substitute the value of b^{2} in equation (4) we get,

\( \dfrac{ 1}{a^2} = \dfrac{1}{16}(1 – \dfrac{9}{b^2})\)

= \(\dfrac{1} {16}(1 – \dfrac{9}{13})\)

= \(\dfrac{ 1}{16}\dfrac{(13-9)}{13}\)

=\( \dfrac{ 1}{16}(\dfrac{4}{13})\)

= \( \dfrac{ 1}{52}\)

a^{2} = 52

Equation of ellipse is \( \dfrac{ x^2}{a^2} – \dfrac{y^2}{b^2} = 1,\)

By substituting the values of a^{2} and b^{2} in above equation we get,

\(\dfrac{ x^2}{52} + \dfrac{y^2}{13} = 1\)

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