Q.Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $4x^2 + 9y^2 = 36$.

Concept: Ellipse Standard Form — rewrite the given equation in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to identify $a$ and $b$, then use $c^2 = a^2 - b^2$ (since $a > b$ here).

Step 1: Normalise the equation

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

So $a^2 = 9$, $b^2 = 4$; hence $a = 3$, $b = 2$. Since $a > b$, the major axis is along the $x$-axis.

Step 2: Find $c$ (distance from centre to each focus)

Step 3: List all required parameters

• Vertices: $(\pm a, 0) = (\pm 3, 0)$
• Foci: $(\pm c, 0) = (\pm \sqrt{5}, 0)$
• Length of major axis: $2a = 6$
• Length of minor axis: $2b = 4$ …

The equation $4x^2 + 9y^2 = 36$ is not yet in the standard form of an ellipse. The standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (or the swapped version if the major axis is vertical). The key idea: divide through by the constant on the right to get 1, then read off $a$ and $b$. The larger denominator tells you which axis is the major axis.

1. Rewrite in standard form. Divide both sides by 36:

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

Hence $a = 3$, $b = 2$.

2. Find $c$ (distance from centre to each focus). For an ellipse, $c^2 = a^2 - b^2$ (when $a > b$).

3. Vertices and foci.

   The centre is at $(0,0)$.

   • Vertices lie on the major axis: $(\pm a, 0) = (\pm 3, 0)$.
   • Foci lie on the major axis inside the ellipse: $(\pm c, 0) = (\pm \sqrt{5}, 0)$.

4. Lengths of axes.

   • Major axis length = $2a = 6$.
   • Minor axis length = $2b = 4$.

5. Eccentricity.

Since $e < 1$, it's an ellipse. The smaller the eccentricity, the more circular the ellipse.

6. Length of latus rectum. …