Let the ellipse E : x² + 9y² = 9 intersect the positive x- and y-axes at the points A and B respecti

Q: Why is the radius of circle $C$ equal to $3$?

The problem states that the major axis of the ellipse $E$ is the diameter of the circle $C$. For the ellipse $\frac{x^2}{9} + \frac{y^2}{1} = 1$, the semi-major axis $a$ is $3$. Therefore, the major axis length is $2a = 6$. Since this length is the diameter of the circle, the radius $r$ is half of that, which is $3$.

Q: When I solve the intersection of the line and the circle, I get two points. How do I know which one is $P$?

Great observation! The line $AB$ intersects the circle at two points. One of these points is $A(3, 0)$, which is the $x$-intercept of the line. The problem defines $P$ as the point where the line meets the circle *other* than $A$ (implied by the context of forming a triangle $APO$). When you solve the quadratic $10y^2 - 18y = 0$, you get $y=0$ (which is point $A$) and $y = \frac{9}{5}$. Thus, $P$ must correspond to $y = \frac{9}{5}$.

Q: Can I use the determinant formula for the area of triangle $APO$?

Absolutely! The determinant formula $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ is a powerful tool. However, since the base $OA$ lies on the $x$-axis, the base-height formula $\frac{1}{2} \times \text{base} \times \text{height}$ is much faster. The base is the length $OA = 3$, and the height is simply the absolute value of the $y$-coordinate of $P$.

Question

Let the ellipse $E : x^2 + 9y^2 = 9$ intersect the positive $x$- and $y$-axes at the points $A$ and $B$ respectively. Let the major axis of $E$ be a diameter of the circle $C$. Let the line passing through $A$ and $B$ meet the circle $C$ at the point $P$. If the area of the triangle with vertices $A, P$ and the origin $O$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, then $m - n$ is equal to

Options:

(A) 16

(B) 15

(C) 17

(D) 18

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of mathematics! Today, we are going to unravel a beautiful problem that bridges the gap between the elegance of conics and the precision of coordinate geometry.

Imagine you are standing on a coordinate plane, looking at an ellipse $E$ defined by $x^2 + 9y^2 = 9$. To truly understand its nature, we must bring it into its standard form.

By dividing the entire equation by $9$, we get:

$$
\frac{x^2}{9} + \frac{y^2}{1} = 1
$$

Here, the semi-major axis $a$ is $3$, and the semi-minor axis $b$ is $1$. This tells us the ellipse is elongated along the $x$-axis, reaching from $-3$ to $3$.

### The Circle as a Container

Now, the problem introduces a circle $C$. We are told the major axis of our ellipse is the diameter of this circle.

Since the major axis length is $2a = 6$, the diameter of our circle is $6$, which means its radius $r$ is $3$. Centered at the origin, the equation for circle $C$ is simply:

$$
x^2 + y^2 = 9
$$

It is fascinating to see how the circle perfectly encloses the ellipse, touching it at the vertices on the $x$-axis. This is the stage where our drama unfolds.

### The Bridge: Line $AB$

Next, we identify points $A$ and $B$. Point $A$ is where the ellipse hits the positive $x$-axis, so $A = (3, 0)$. Point $B$ is where it hits the positive $y$-axis, so $B = (0, 1)$.

We draw a line connecting these two points. Using the intercept form, the equation of this line is:

$$
\frac{x}{3} + \frac{y}{1} = 1
$$

To make our lives easier, let's rewrite this as $x = 3 - 3y$. This simple linear relationship is the key to unlocking the coordinates of point $P$, the intersection of our line and the circle.

### The Algebraic Dance

Now, let's find $P$. We substitute our line equation into the circle's equation:

$$
(3 - 3y)^2 + y^2 = 9
$$

Expanding the squared term, we get $9 - 18y + 9y^2 + y^2 = 9$. Notice the magic? The $9$s on both sides cancel out, leaving us with:

$$
10y^2 - 18y = 0
$$

Factoring this, we get $2y(5y - 9) = 0$. This gives us two solutions: $y = 0$ and $y = \frac{9}{5}$.

The $y=0$ solution corresponds to point $A(3, 0)$. The other solution, $y = \frac{9}{5}$, must be our point $P$. Plugging this back into our line equation, we find:

$$
x = 3 - 3\left(\frac{9}{5}ight) = 3 - \frac{27}{5} = -\frac{12}{5}
$$

So, $P = \left(-\frac{12}{5}, \frac{9}{5}ight)$.

### The Final Victory

We are almost there! We need the area of triangle $APO$, with vertices $O(0, 0)$, $A(3, 0)$, and $P\left(-\frac{12}{5}, \frac{9}{5}ight)$.

Since the base $OA$ lies on the $x$-axis, its length is $3$. The height of the triangle is the perpendicular distance from $P$ to the $x$-axis, which is simply the $y$-coordinate of $P$, or $\frac{9}{5}$.

The area is:

$$
	ext{Area} = \frac{1}{2} 	imes 3 	imes \frac{9}{5} = \frac{27}{10}
$$

The problem tells us this area is $\frac{m}{n}$, where $m$ and $n$ are coprime. Thus, $m = 27$ and $n = 10$.

Finally, **$m - n = 27 - 10 = 17$**. We have successfully navigated the geometry and arrived at our destination.

Select Answer:

Visualized Solution

Standard Form of Ellipse $E$

Coordinates of Points $A$ and $B$

Equation of Circle $C$

Equation of Line $A B$

Intersection of Line and Circle

Solving the Quadratic Equation

Coordinates of Point $P$

Forming Triangle $A P O$

Calculating Area of Triangle $A P O$

Final Answer: $m - n$

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

Analyzing the Setup

The Circle as a Container

The Bridge

The Algebraic Dance

The Final Victory

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