Question

The foci of the ellipse $4x^2 + 9y^2 = 1$ are:

• A. $\left( \pm \frac{\sqrt{3}}{2}, 0 \right)$
• B. $\left( \pm \frac{\sqrt{5}}{2}, 0 \right)$
• C. $\left( \pm \frac{\sqrt{5}}{3}, 0 \right)$
• D. $\left( \pm \frac{\sqrt{5}}{6}, 0 \right)$
• E. $\left( \pm \frac{\sqrt{5}}{4}, 0 \right)$

Hint:

When the ellipse equation has coefficients like $4$ and $9$, the denominators in standard form are their reciprocals. Always check which reciprocal is larger to determine if the ellipse is horizontal or vertical.

Answer 1

The Correct Option is D

Concept: The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are located at $(\pm ae, 0)$ if $a > b$, where $e$ is the eccentricity.
• Relation for $c$ (distance from center to focus): $c^2 = a^2 - b^2$.
• Foci coordinates: $(\pm c, 0)$.

Step 1: Convert the equation to standard form.
Divide the given equation $4x^2 + 9y^2 = 1$ to express it in terms of denominators: \[ \frac{x^2}{1/4} + \frac{y^2}{1/9} = 1 \] Comparing with the standard form, we find $a^2 = \frac{1}{4}$ and $b^2 = \frac{1}{9}$. Since $\frac{1}{4} > \frac{1}{9}$, the major axis is along the x-axis.

Step 2: Calculate the distance to the foci ($c$).
Using the formula $c^2 = a^2 - b^2$: \[ c^2 = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \] Taking the square root: \[ c = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \]

Step 3: Determine the coordinates.
The foci are $(\pm c, 0)$: \[ \text{Foci} = \left( \pm \frac{\sqrt{5}}{6}, 0 \right) \]