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(A) Since the major axis is on the $x-$ axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(1)$.Given th

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$\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$

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(A) Since the major axis is on the $x-$ axis,the equation of the ellipse is of the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(1)$.Given that the ellipse passes through $(4, 3)$ and $(6, 2)$,we have:$\frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$ $(2)$$\frac{36}{a^{2}} + \frac{4}{b^{2}} = 1$ $(3)$Let $u = \frac{1}{a^{2}}$ and $v = \frac{1}{b^{2}}$. The equations become:$16u + 9v = 1$$36u + 4v = 1$Multiplying the first by $4$ and the second by $9$:$64u + 36v = 4$$324u + 36v = 9$Subtracting the equations: $260u = 5 \implies u = \frac{5}{260} = \frac{1}{52}$.Thus,$a^{2} = 52$.Substituting $u$ into $36u + 4v = 1$:$36(\frac{1}{52}) + 4v = 1 \implies \frac{9}{13} + 4v = 1 \implies 4v = 1 - \frac{9}{13} = \frac{4}{13} \implies v = \frac{1}{13}$.Thus,$b^{2} = 13$.The equation is $\frac{x^{2}}{52} + \frac{y^{2}}{13} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4, 3)$ and $(6, 2)$.

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