Example 13 Find the equation' of the ellipse, with major axis along the \( x \)-axis ax passing through the points \( (4,3) \) and \( (-1,4) \).

Let the equation of ellipse be

\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 ....(1)

Ellipse is passing through points (4,3) & (-1,4) and major axis be x-axis.

∴ \(\frac{16}{a^2}\) + \(\frac{9}{b^2}\) = 1 ....(2)

\(\frac{1}{a^2}\) + \(\frac{16}{b^2}\) = 1 ....(3)

Multiplying equation (3) by 16,we get

\(\frac{16}{a^2}\) + \(\frac{256}{b^2}\) = 16 ....(4)

Subtract equation (2) from (4),we get

\(\frac{256-9}{b^2}\) = 16 - 1

⇒ \(\frac{247}{b^2}\) = 15

⇒ b² = \(\frac{247}{15}\)

Put b² = \(\frac{247}{15}\) in equation (3), we get

\(\frac{1}{a^2}\)+\(\frac{16\times 15}{247}\) = 1

⇒ \(\frac{1}{a^2}\) = 1 - \(\frac{240}{247}\)

= \(\frac{7}{247}\)

⇒ a² = \(\frac{247}{7}\).

By putting a² = \(\frac{247}{7}\) & b² = \(\frac{247}{15}\) in equation (1) we get,

\(\frac{x^2}{\frac{247}{7}}\) + \(\frac{y^2}{\frac{247}{15}}\) = 1

⇒ \(\frac{7x^2}{247}\) + \(\frac{15y^2}{247}\) = 1

Which is required equation of ellipse.