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 which vertices A, P and the origin O is m/n , where m and n are coprime, then m – n is equal to 

• A. 15
• B. 16
• C. 17
• D. 18

Answer 1

The Correct Option is C

Given: The circle is defined by the equation:

\( x^2 + y^2 = 9 \)

Step 1: Equation of Line AB or AP: The equation of line AB is:

\( \frac{x}{3} + \frac{y}{1} = 1 \)

\( x + 3y = 3 \Rightarrow x = 3 - 3y \)

Step 2: Intersection Point of Line AP and the Circle: Substitute \( x = 3 - 3y \) into \( x^2 + y^2 = 9 \):

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

\( 9(1 + y^2 - 2y) + y^2 = 9 \)

\( 10y^2 - 18y = 0 \)

\( y(10y - 18) = 0 \Rightarrow y = \frac{9}{5} \)

Substitute \( y = \frac{9}{5} \) into \( x = 3 - 3y \):

\( x = 3 \left(1 - \frac{9}{5} \right) = 3 \left( -\frac{4}{5} \right) = -\frac{12}{5} \)

Thus, the intersection point is:

\( P \left( -\frac{12}{5}, \frac{9}{5} \right) \)

Step 3: Area of △AOP: The area of △AOP is given by:

Area \(= \frac{1}{2} \times \text{Base (OA)} \times \text{Height}\)

Here, Base (OA) = 3 and Height =\(\frac{9}{5}\):

Area \(= \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}\)

Step 4: Express the Area as \(\frac{m}{n}\):

Area \(= \frac{27}{10}, \quad m = 27, \quad n = 10\)

Step 5: Calculate \(m-n\):

\( m - n = 27 - 10 = 17 \)

Final Answer: \( m - n = 17 \), which corresponds to Option C.