Q.Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x$-axis and passes through the points $(4, 3)$ and $(6, 2)$.

Concept: Locus of a point satisfying the standard ellipse equation with major axis along the $x$-axis.

The standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$.

Since the ellipse passes through $(4, 3)$:

Since it passes through $(6, 2)$:

Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The system becomes:

Multiply equation (1) by 4 and equation (2) by 9:

$324u + 36v = 9$ …

Understanding the Problem

When the major axis lies on the $x$-axis, the ellipse has its longer dimension horizontal. The standard form is

where $a > b > 0$. Here $a$ is the semi-major axis (half the length along $x$) and $b$ is the semi-minor axis (half the length along $y$).

The key insight: any point $(x, y)$ on the ellipse must satisfy this equation. Since we're given two points that lie on the ellipse, we can substitute their coordinates to generate two equations in the two unknowns $a^2$ and $b^2$.

Solution

1. Set up the equation for point $(4, 3)$

Substituting into the standard form:

2. Set up the equation for point $(6, 2)$

Similarly:

3. Solve the system

Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$ to simplify. The system becomes:

Multiply equation (i) by 4 and equation (ii) by 9:

Subtract (iii) from (iv):

Therefore $a^2 = 52$.

4. Find $b^2$ …