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
- 15
- 16
- 17
- 18
The Correct Option is C
Solution and Explanation
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.
Top Questions on Triangles
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
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If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
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