Find eccentricity of ellipse 4x²+9y²=36

Question

Find the eccentricity of the ellipse $4x^2 + 9y^2 = 36$ . Also find the foci and length of the major and minor axes.

Solution — Step by Step

The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Divide both sides of $4x^2 + 9y^2 = 36$ by 36:

\frac{4x^2}{36} + \frac{9y^2}{36} = 1

\frac{x^2}{9} + \frac{y^2}{4} = 1

So: $a^2 = 9$ , $b^2 = 4$ , giving $a = 3$ , $b = 2$ .

Since $a^2 = 9 > b^2 = 4$ , we have $a > b$ , and the major axis is along the x-axis.

For an ellipse with major axis along x-axis:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b

The major axis has semi-length $a = 3$ ; minor axis has semi-length $b = 2$ .

For an ellipse: $c^2 = a^2 - b^2$

c^2 = 9 - 4 = 5

c = \sqrt{5}

$c$ is the distance from the centre to each focus.

e = \frac{c}{a} = \frac{\sqrt{5}}{3}

Eccentricity $e = \frac{\sqrt{5}}{3} \approx 0.745$

For an ellipse, $0 < e < 1$ . Our answer satisfies this (as $\sqrt{5} \approx 2.236 < 3$ ). ✓

Foci: Located at $(\pm c, 0) = (\pm\sqrt{5}, 0)$

Length of major axis $= 2a = 6$

Length of minor axis $= 2b = 4$

Summary:

$e = \sqrt{5}/3$
Foci: $(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$
Major axis length: 6 (along x-axis)
Minor axis length: 4 (along y-axis)

Why This Works

Eccentricity measures how “stretched” an ellipse is. A circle has $e = 0$ (perfectly round, both foci at centre). As $e \to 1$ , the ellipse becomes more elongated. The relationship $c^2 = a^2 - b^2$ comes from the definition of an ellipse (sum of distances from two foci = constant $= 2a$ ) combined with the geometry of the foci at $(\pm c, 0)$ .

Key formula chain to memorise: $e = c/a$ , $c^2 = a^2 - b^2$ (ellipse), $c^2 = a^2 + b^2$ (hyperbola). For a circle, $a = b$ so $c = 0$ and $e = 0$ . For a parabola, $e = 1$ exactly. These four cases (circle, ellipse, parabola, hyperbola) are the four conic sections, and eccentricity uniquely identifies each.

Alternative — Using Latus Rectum

Length of latus rectum $= \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$

This is the chord through a focus perpendicular to the major axis — a common sub-question in JEE.

Common Mistake

The most common error is writing $c^2 = a^2 + b^2$ (the Pythagorean theorem for right triangles, or the formula for a hyperbola). For an ellipse, $c^2 = a^2 - b^2$ — the foci are “inside” the ellipse, so $c < a$ , and we subtract. For a hyperbola, the relationship is $c^2 = a^2 + b^2$ — the foci are “outside,” and $c > a$ . Get this distinction wrong, and every ellipse problem gives wrong foci and wrong eccentricity.

Question

Solution — Step by Step

Why This Works

Alternative — Using Latus Rectum

Common Mistake

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