Q.Find the equation for the ellipse that satisfies the given conditions: Centre at $(0, 0)$, major axis on the $y$-axis and passes through the points $(3, 2)$ and $(1, 6)$.

Concept: Locus of Point — The ellipse is defined by its standard equation, and the given points must satisfy it.

Since the major axis is on the $y$-axis and centre is at $(0,0)$, the standard form is:

Substitute $(3, 2)$:

Substitute $(1, 6)$:

Solve these two equations. Multiply the second by 9 and subtract from the first:

$\left(\frac{9}{b^2} + \frac{4}{a^2}\right) - \left(\frac{9}{b^2} + \frac{324}{a^2}\right) = 1 - 9$ …

The key idea: when the major axis lies on the $y$-axis, the ellipse is "taller" than it is wide. That means the larger denominator goes under $y^2$, not $x^2$. Many students instinctively put the larger number under $x^2$ because they're used to horizontal ellipses — that's the classic trap here.

So we start with the standard form for a vertical ellipse centered at the origin:

Here $a$ is the semi-major axis (vertical), $b$ is the semi-minor axis (horizontal). We don't know $a^2$ or $b^2$ yet — but we have two points the ellipse passes through. Each point gives us one equation.

1. Substitute $(3, 2)$ into the equation:

2. Substitute $(1, 6)$ into the equation:

Now we have two equations in the unknowns $\frac{1}{b^2}$ and $\frac{1}{a^2}$. Let’s set:

Then the system becomes:

3. Solve for $u$ and $v$. From Equation 2: $u = 1 - 36v$. Substitute into Equation 1:

So $a^2 = \frac{1}{v} = 40$.

4. Find $u$: $u = 1 - 36 \cdot \frac{1}{40} = 1 - \frac{36}{40} = 1 - \frac{9}{10} = \frac{1}{10}$. …