Mathematical Toolkit: Essential Concepts and Formulas

Before embarking on the solution, let's establish the fundamental mathematical tools we'll be employing. A strong grasp of these concepts is crucial for navigating problems involving conic sections, lines, and areas in coordinate geometry.

1. Standard Form of an Ellipse: An ellipse centered at the origin $(0,0)$ is typically represented by the equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

   • If $a > b$, the major axis lies along the x-axis, with length $2a$. The vertices are at $(\pm a, 0)$.
   • If $b > a$, the major axis lies along the y-axis, with length $2b$. The vertices are at $(0, \pm b)$.
   • To find x-intercepts, set $y=0$. To find y-intercepts, set $x=0$.

2. Equation of a Line (Intercept Form): When a line intersects the x-axis at $(x_0, 0)$ and the y-axis at $(0, y_0)$, its equation can be efficiently written as: $\frac{x}{x_0} + \frac{y}{y_0} = 1$ This form is particularly useful when the intercepts are known or easily determined.

3. Equation of a Circle: A circle with its center at the origin $(0,0)$ and a radius $r$ has the equation: $x^2 + y^2 = r^2$

4. Intersection of Geometric Shapes: To find the points where a line and a circle intersect, we solve their equations simultaneously. This usually involves substituting an expression for one variable from the linear equation into the quadratic equation of the circle, leading to a quadratic equation in a single variable.

5. Area of a Triangle: For a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and the origin $(0,0)$, the area can be calculated using the formula: $\text{Area} = \frac{1}{2} |x_1 y_2 - x_2 y_1|$ Alternatively, if one side of the triangle lies along a coordinate axis, we can use the familiar formula: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$ where the height is the perpendicular distance from the third vertex to the base.

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Step-by-Step Elaborated Solution:

1. Analyze the Ellipse $E$ and Determine Intercepts A and B

The given equation of the ellipse is $E: x^2 + 9y^2 = 9$. Our first objective is to understand the ellipse's dimensions and orientation, and specifically locate the points where it crosses the positive x and y-axes.

• Transforming to Standard Form: To clearly identify the semi-axes, we convert the given equation into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We achieve this by dividing the entire equation by 9: $\frac{x^2}{9} + \frac{9y^2}{9} = \frac{9}{9}$ This simplifies to: $\frac{x^2}{3^2} + \frac{y^2}{1^2} = 1$ By comparing this with the standard form, we deduce:

  • $a^2 = 9 \Rightarrow a = 3$ (The semi-major axis length, as $a$ must be positive).
  • $b^2 = 1 \Rightarrow b = 1$ (The semi-minor axis length, as $b$ must be positive).

• Identifying the Major Axis and Intercepts: Since $a=3$ and $b=1$, we have $a > b$. This indicates that the major axis of the ellipse lies along the x-axis. Its total length is $2a = 2(3) = 6$. The vertices along the major axis are $(\pm 3, 0)$.

  Now, we find the specific points A and B as described in the problem:

  • Point A (Positive x-intercept): We set $y=0$ in the ellipse equation: $x^2 + 9(0)^2 = 9 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. The problem specifies A as the intercept on the positive x-axis, so we choose $x=3$. Thus, $A = (3, 0)$.
  • Point B (Positive y-intercept): We set $x=0$ in the ellipse equation: $(0)^2 + 9y^2 = 9 \Rightarrow 9y^2 = 9 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$. Similarly, B is on the positive y-axis, so we choose $y=1$. Thus, $B = (0, 1)$.

  Tip for JEE: Always read keywords carefully. "Positive x and y-axes" is crucial for selecting the correct intercepts from the $\pm$ possibilities.

2. Determine the Equation of Line AB

With the coordinates of $A(3, 0)$ and $B(0, 1)$ established, our next step is to find the equation of the line passing through these two points.

• Using the Intercept Form: Since A is the x-intercept ($x_0 = 3$) and B is the y-intercept ($y_0 = 1$), the intercept form of a linear equation is the most direct and efficient method: $\frac{x}{x_0} + \frac{y}{y_0} = 1$ Substituting the values: $\frac{x}{3} + \frac{y}{1} = 1$
• Converting to General Form: To simplify further calculations, especially for substitution later, it's often helpful to clear the denominators and express the equation in the general form $Ax+By=C$. Multiply the entire equation by the least common multiple of the denominators (which is 3): $3 \left(\frac{x}{3}\right) + 3 \left(\frac{y}{1}\right) = 3(1)$ $\Rightarrow x + 3y = 3 \quad \text{.........(Equation of line AB)}$

3. Determine the Equation of Circle C

The problem states that the major axis of ellipse E is a diameter of circle C. We've already determined the properties of the major axis.

• Finding the Diameter and Radius: From Step 1, the major axis of ellipse E lies along the x-axis and has a length of $2a = 6$. Therefore, the diameter of circle C is 6 units. The radius of circle C is $r = \frac{\text{diameter}}{2} = \frac{6}{2} = 3$ units.

• Determining the Center of the Circle: Since the major axis of the ellipse (which is centered at the origin) is given as a diameter of the circle, it logically follows that the center of circle C must also be the origin $(0,0)$. This is the most natural interpretation of the statement; otherwise, the major axis would only be a chord, not a diameter, unless the circle was specifically constructed to have its center on the major axis.

• Formulating the Equation of Circle C: Using the standard equation for a circle centered at the origin $(0,0)$ with radius $r$: $x^2 + y^2 = r^2$. Substituting $r=3$: $x^2 + y^2 = 3^2$ $\Rightarrow x^2 + y^2 = 9 \quad \text{.........(Equation of circle C)}$

4. Find the Point of Intersection P

The line AB intersects the circle C at point P. We need to find the coordinates of P by solving the equations of line AB and circle C simultaneously.

We have:

1. Line AB: $x + 3y = 3$
2. Circle C: $x^2 + y^2 = 9$

• Strategy: Substitution Method: We will express one variable from the linear equation (line AB) in terms of the other and substitute it into the quadratic equation (circle C). This allows us to solve for one variable first. From the equation of line AB, it's easiest to express $x$ in terms of $y$: $x = 3 - 3y$

• Substituting into the Circle Equation: Now, substitute this expression for $x$ into the equation of circle C: $(3 - 3y)^2 + y^2 = 9$

• Solving the Quadratic Equation: Expand the squared term using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$: $(3^2 - 2 \cdot 3 \cdot 3y + (3y)^2) + y^2 = 9$ $(9 - 18y + 9y^2) + y^2 = 9$ Combine the $y^2$ terms: $10y^2 - 18y + 9 = 9$ Subtract 9 from both sides to simplify: $10y^2 - 18y = 0$ Factor out the common term $2y$: $2y(5y - 9) = 0$ This yields two possible values for $y$:

  • $2y = 0 \Rightarrow y = 0$
  • $5y - 9 = 0 \Rightarrow 5y = 9 \Rightarrow y = \frac{9}{5}$

• Finding Corresponding x-coordinates: We use the relation $x = 3 - 3y$ to find the $x$-coordinates for each $y$-value:

  • For $y = 0$: $x = 3 - 3(0) = 3$. This gives the point $(3, 0)$. This is precisely point A, which we already know is an intersection point. This serves as a useful check of our calculations.
  • For $y = \frac{9}{5}$: $x = 3 - 3\left(\frac{9}{5}\right) = 3 - \frac{27}{5}$. To combine these terms, find a common denominator: $x = \frac{15}{5} - \frac{27}{5} = \frac{15 - 27}{5} = -\frac{12}{5}$. This gives the point $\left(-\frac{12}{5}, \frac{9}{5}\right)$.

  Since A is one intersection point, the other point must be P. Therefore, the coordinates of point $P$ are $\left(-\frac{12}{5}, \frac{9}{5}\right)$.

  Common Mistake: When solving the quadratic, remember that you are looking for two intersection points. One will correspond to point A, and the other to point P. Don't stop after finding the first one if the problem asks for the "other" point.

5. Calculate the Area of Triangle APO

We need to find the area of the triangle with vertices $A(3, 0)$, $P\left(-\frac{12}{5}, \frac{9}{5}\right)$, and the origin $O(0, 0)$.

• Method 1: Using Base and Height (Most Efficient Here) Notice that vertex A $(3,0)$ and the origin O $(0,0)$ both