The ellipse x² + 4y² = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn

Q: Wait, why do we assume the rectangle vertices are $(2, 1)$? Couldn't they be somewhere else?

That is a fantastic question! The problem states the rectangle is 'aligned with the coordinate axes' and 'inscribed' in the inner ellipse. This means the sides of the rectangle are parallel to the axes and must touch the ellipse at its extreme points—the vertices. For the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$, the extreme points are $(\pm 2, 0)$ and $(0, \pm 1)$. The rectangle connecting these points must have corners at $(\pm 2, \pm 1)$. If the rectangle were anywhere else, it wouldn't be 'inscribed' in the strict sense of touching the ellipse at its boundaries.

Q: Why does the point $(4, 0)$ tell us $A^2$ directly? Is it always that simple?

It is that simple because of the coordinate geometry! Think about the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. If a point lies on the x-axis, its y-coordinate is $0$. Substituting $(4, 0)$ gives $\frac{4^2}{A^2} + \frac{0^2}{B^2} = 1$, which simplifies to $\frac{16}{A^2} = 1$. This forces $A^2 = 16$. Whenever you are given an intercept on the axis, you are essentially being handed the semi-axis length on a silver platter!

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure coordinate geometry problem. While physics often uses ellipses (like planetary orbits), here we are dealing with the static properties of conic sections. We rely on the algebraic definition of the ellipse. Stick to the coordinate definitions—they are the most powerful tools in your arsenal for this specific challenge.

Question

The ellipse $x^2 + 4y^2 = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4, 0)$. Then the equation of the ellipse is:

Options:

(A) $x^2 + 12y^2 = 16$

(B) $4x^2 + 48y^2 = 48$

(C) $4x^2 + 64y^2 = 48$

(D) $x^2 + 16y^2 = 16$

Sigma Percentile · Accepted Answer

### Analyzing the Inner Ellipse

We begin with the inner ellipse, defined by the equation $x^2 + 4y^2 = 4$. To truly see its nature, we must bring it into its standard form.

We divide the entire equation by $4$, yielding:

$$\frac{x^2}{4} + \frac{y^2}{1} = 1$$

This equation reveals that the semi-major axis $a$ is $2$, and the semi-minor axis $b$ is $1$. The ellipse stretches from $-2$ to $2$ along the $x$-axis and from $-1$ to $1$ along the $y$-axis. It is a perfect, compact oval centered at the origin.

### The Bridge of the Rectangle

The problem introduces a rectangle 'inscribed' in this inner ellipse. In the language of geometry, this means the rectangle is tightly packed, with its sides touching the ellipse.

Because the ellipse is aligned with the axes, the rectangle's corners must be the points where the ellipse reaches its maximum extent. These are the points $(\pm 2, \pm 1)$. This rectangle acts as our bridge and the anchor point for the outer, larger ellipse.

### The Outer Ellipse and the Mystery Point

We define the outer ellipse with the general equation:

$$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$

We have two unknowns here: $A^2$ and $B^2$. The problem provides two keys: the outer ellipse passes through the point $(4, 0)$ and it circumscribes our rectangle, meaning it must pass through the rectangle's corner, $P(2, 1)$.

### Unlocking the Constants

Let us tackle the point $(4, 0)$ first. Substituting $x = 4$ and $y = 0$ into our general equation, we get:

$$\frac{4^2}{A^2} + \frac{0^2}{B^2} = 1 \Rightarrow \frac{16}{A^2} = 1 \Rightarrow A^2 = 16$$

Now, we turn to the corner point $P(2, 1)$. Since the outer ellipse must pass through this point, we substitute $x = 2$, $y = 1$, and our known $A^2 = 16$ into the equation:

$$\frac{2^2}{16} + \frac{1^2}{B^2} = 1$$

This simplifies to:

$$\frac{4}{16} + \frac{1}{B^2} = 1 \Rightarrow \frac{1}{4} + \frac{1}{B^2} = 1$$

Subtracting $\frac{1}{4}$ from both sides, we find:

$$\frac{1}{B^2} = \frac{3}{4} \Rightarrow B^2 = \frac{4}{3}$$

### The Final Synthesis

We have our constants and our shape. Let us assemble the final equation:

$$\frac{x^2}{16} + \frac{y^2}{\frac{4}{3}} = 1$$

To make this look elegant, we simplify the second term. Dividing by a fraction is equivalent to multiplying by its reciprocal, so $\frac{y^2}{4/3}$ becomes $\frac{3y^2}{4}$. Our equation is now:

$$\frac{x^2}{16} + \frac{3y^2}{4} = 1$$

To clear the denominators, we multiply the entire equation by $16$:

$$x^2 + 16 \cdot \left( \frac{3y^2}{4} \right) = 16$$

The final equation of the outer ellipse is:

**$x^2 + 12y^2 = 16$**

Animated Solution for Mathematics - Conic Sections: The ellipse $x^{2} + 4 y^{2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4, 0)$ . Then the equation of the ellipse is:

Select Answer:

Visualized Solution

Inner Ellipse Equation

Inscribed Rectangle

Outer Ellipse Setup

Given Point on Outer Ellipse

Substituting $(4, 0)$

Calculating $A^{2}$

Substituting Vertex $P (2, 1)$

Simplifying the Equation

Calculating $B^{2}$

Constructing the Final Equation

Rearranging the Terms

Final Standard Form

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Inner Ellipse

The Bridge of the Rectangle

The Outer Ellipse and the Mystery Point

Unlocking the Constants

The Final Synthesis

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