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(A) The standard form of the ellipse with the major axis along the $x-$axis is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.Since the points $(4, 3

Vedclass · Accepted Answer

$7x^{2} + 15y^{2} = 247$

Vedclass · Answer

(A) The standard form of the ellipse with the major axis along the $x-$axis is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.Since the points $(4, 3)$ and $(-1, 4)$ lie on the ellipse,we have:$\frac{16}{a^{2}} + \frac{9}{b^{2}} = 1$  --- $(1)$$\frac{1}{a^{2}} + \frac{16}{b^{2}} = 1$  --- $(2)$Let $u = \frac{1}{a^{2}}$ and $v = \frac{1}{b^{2}}$. The equations become:$16u + 9v = 1$  --- $(3)$$u + 16v = 1$  --- $(4)$Multiplying $(4)$ by $16$,we get $16u + 256v = 16$. Subtracting $(3)$ from this:$247v = 15 \implies v = \frac{15}{247} \implies b^{2} = \frac{247}{15}$.Substituting $v$ into $(4)$:$u + 16(\frac{15}{247}) = 1 \implies u = 1 - \frac{240}{247} = \frac{7}{247} \implies a^{2} = \frac{247}{7}$.Substituting $a^{2}$ and $b^{2}$ into the standard form:$\frac{x^{2}}{247/7} + \frac{y^{2}}{247/15} = 1 \implies \frac{7x^{2}}{247} + \frac{15y^{2}}{247} = 1 \implies 7x^{2} + 15y^{2} = 247$.

Find the equation of the ellipse with the major axis along the $x-$axis and passing through the points $(4, 3)$ and $(-1, 4)$.

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