Let C be the circle of minimum area enclosing the ellipse E : x²/a² + y²/b² = 1 with eccentricity 1/

Q: Wait, why is the minimum enclosing circle the auxiliary circle?

Great question! The auxiliary circle of an ellipse is the circle with the same center and a radius equal to the semi-major axis $a$. Any circle smaller than this would fail to cover the vertices of the ellipse at $(\pm a, 0)$. Any circle larger would have a greater area. Thus, the auxiliary circle is the unique circle of minimum area that encloses the ellipse.

Q: The problem says the base length is 29, but that seems impossible. What do I do?

You have a sharp eye! In competitive exams, typos can occasionally occur. Given the context of the ellipse with $a=4$, a base length of 29 is geometrically inconsistent with the problem's constraints. Based on the standard geometry of such problems, it is almost certain that '29' is a typo for $2a$, which is $2 \times 4 = 8$. Proceed with the base length as 8.

Q: Why is the height $y + 2\sqrt{3}$?

The height of a triangle is the perpendicular distance from the vertex $P(x, y)$ to the base line. Our base line is $y = -2\sqrt{3}$. The vertical distance from any point $y$ to the line $y = -2\sqrt{3}$ is simply $y - (-2\sqrt{3})$, which simplifies to $y + 2\sqrt{3}$. To maximize this, we simply need to maximize $y$.

Question

Let $C$ be the circle of minimum area enclosing the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $\frac{1}{2}$ and foci $(\pm 2, 0)$. Let $PQR$ be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $QR$ of length 29 is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $PQR$ is :

Options:

(A) $6(3 + \sqrt{2})$

(B) $8(3 + \sqrt{2})$

(C) $62 + \sqrt{3}$

(D) $8(2 + \sqrt{3})$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Ellipse

Consider the ellipse $E$ defined by the equation:

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

We are given the foci at $(\pm 2, 0)$, which implies the focal distance $ae = 2$. Given the eccentricity $e = \frac{1}{2}$, we can determine the semi-major axis $a$:

$$
a = \frac{2}{e} = \frac{2}{1/2} = 4
$$

Next, we determine the semi-minor axis $b$ using the fundamental relationship $b^2 = a^2(1 - e^2)$:

$$
b^2 = 16 \left(1 - \frac{1}{4}ight) = 16 \left(\frac{3}{4}ight) = 12
$$

Thus, the semi-minor axis is $b = 2\sqrt{3}$. We have now fully defined the geometric profile of the ellipse.

### The Guardian Circle: The Auxiliary Circle

We consider the circle $C$ of minimum area that encloses the ellipse. For the circle to be minimal, it must be the auxiliary circle, which touches the ellipse at the vertices of the major axis.

The radius of this circle is equal to the semi-major axis $a = 4$. Therefore, the equation of the circle $C$ is:

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

This circle serves as the symmetric boundary for our optimization problem.

### The Triangle and the Optimization

We define the triangle $PQR$ such that the side $QR$ is parallel to the major axis and passes through the intersection of the ellipse with the negative $y$-axis. Since the ellipse intersects the $y$-axis at $(0, -b)$, the line containing $QR$ is:

$$
y = -2\sqrt{3}
$$

The length of the base $QR$ is equal to the length of the major axis, which is $2a = 8$. To maximize the area of $\Delta PQR$ with a fixed base, we must maximize the vertical height $h$ from a point $P(x, y)$ on the circle to the line $y = -2\sqrt{3}$.

The height $h$ is given by:

$$
h = y - (-2\sqrt{3}) = y + 2\sqrt{3}
$$

### The Final Calculation

To maximize $h$, we must choose the point $P$ on the circle $x^2 + y^2 = 16$ with the largest possible $y$-coordinate. The maximum value for $y$ on this circle is $y = 4$.

Substituting this into our height expression, we find the maximum height:

$$
h_{max} = 4 + 2\sqrt{3}
$$

Finally, we calculate the maximum area of the triangle:

$$
	ext{Area} = \frac{1}{2} 	imes 	ext{Base} 	imes h_{max}
$$

$$
	ext{Area} = \frac{1}{2} 	imes 8 	imes (4 + 2\sqrt{3}) = 4(4 + 2\sqrt{3}) = 16 + 8\sqrt{3}
$$

The maximum area of the triangle is **$8(2 + \sqrt{3})$**.

Select Answer:

Visualized Solution

Analyze the Ellipse $E$

Minimum Enclosing Circle $C$

Locate the Base $QR$

Maximize Height of $Δ P QR$

Calculate Maximum Area

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

Analyzing the Geometry of the Ellipse

The Guardian Circle

The Triangle and the Optimization

The Final Calculation

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