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 $36x^2 + 4y^2 = 144$.

Concept: Ellipse Standard Form – The given equation must be rewritten as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to identify the major axis.

Step 1: Normalise the equation

Divide $36x^2 + 4y^2 = 144$ by $144$:

Step 2: Identify $a$ and $b$

Here $a^2 = 36$ (denominator under $y^2$) and $b^2 = 4$ (under $x^2$). Since $a > b$, the major axis is vertical.

So $a = 6$, $b = 2$.

Step 3: Compute required parameters

• $c = \sqrt{a^2 - b^2} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$
• Vertices: $(0, \pm a) = (0, \pm 6)$
• Foci: $(0, \pm c) = (0, \pm 4\sqrt{2})$
• Length of major axis: $2a = 12$
• Length of minor axis: $2b = 4$ …

The equation $36x^2 + 4y^2 = 144$ is not yet in the standard form for an ellipse. We need to divide through by $144$ to get $1$ on the right-hand side. That gives:

Now compare with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for an ellipse centred at the origin. Here $a^2 = 4$ and $b^2 = 36$. Since $b^2 > a^2$, the major axis is along the $y$-axis (vertical). For a vertical ellipse, we usually write $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ with $a > b$, but the convention is: the larger denominator gives the major axis. So here $a^2 = 36$ (semi-major axis) and $b^2 = 4$ (semi-minor axis). Thus $a = 6$, $b = 2$.

Now we find $c$, the distance from the centre to each focus, using $c^2 = a^2 - b^2$ for an ellipse. So:

Since the major axis is vertical, the foci lie on the $y$-axis at $(0, \pm c) = (0, \pm 4\sqrt{2})$.

The vertices are the endpoints of the major axis, at $(0, \pm a) = (0, \pm 6)$.

The length of the major axis is $2a = 12$, and the length of the minor axis is $2b = 4$.

Eccentricity $e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$.

The length of the latus rectum for an ellipse is $\frac{2b^2}{a}$. Here: …