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Worked Examples · Example 13

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

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The ellipse is x2247/7+y2247/15=1\dfrac{x^2}{247/7} + \dfrac{y^2}{247/15} = 1, i.e. 7x2+15y2=2477x^2 + 15y^2 = 247.

Solution

With the major axis along the xx-axis, take x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1. Let u=1a2, v=1b2u = \dfrac{1}{a^2},\ v = \dfrac{1}{b^2}.

Substituting the two points:

16u+9v=1(from (4,3))16u + 9v = 1 \qquad (\text{from } (4,3))

u+16v=1(from (−1,4))u + 16v = 1 \qquad (\text{from } (-1,4))

From the second equation u=1−16vu = 1 - 16v. Substituting into the first:

16(1−16v)+9v=1  ⇒  16−247v=1  ⇒  v=1524716(1 - 16v) + 9v = 1 \;\Rightarrow\; 16 - 247v = 1 \;\Rightarrow\; v = \frac{15}{247}

Then u=1−16⋅15247=7247u = 1 - 16\cdot\dfrac{15}{247} = \dfrac{7}{247}. Hence

a2=1u=2477,b2=1v=24715.a^2 = \frac{1}{u} = \frac{247}{7}, \qquad b^2 = \frac{1}{v} = \frac{247}{15}. …

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