The line passing through the extremity A of major axis and extremity B of the minor axes of the ellipse `9x^2 + 16y^2 = 144` meets the circle `x^2+ y^2 =16` at the point P. Then the area of the triangle OAP, O being the origin (in square units) is

Options

A. 96/25
B. 192/25
C. 48/25
D. 96/50

Answer

To solve the problem, we need to find the area of triangle OAP, where O is the origin, A is the extremity of the major axis of the ellipse, and P is the intersection point of the line through A and B (extremity of the minor axis) with the circle.

### Step 1: Identify the ellipse parameters
The given ellipse is \(9x^2 + 16y^2 = 144\).

To convert this into standard form, we divide by 144:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
From this, we can identify:
- \(a^2 = 16\) (so \(a = 4\))
- \(b^2 = 9\) (so \(b = 3\))

### Step 2: Find the coordinates of points A and B
The extremity A of the major axis (along the x-axis) is at:
\[ A(4, 0) \]
The extremity B of the minor axis (along the y-axis) is at:
\[ B(0, 3) \]

### Step 3: Find the equation of line AB
The slope of line AB can be calculated as:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 4} = -\frac{3}{4} \]
Using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(A(4, 0)\):
\[ y - 0 = -\frac{3}{4}(x - 4) \]
This simplifies to:
\[ y = -\frac{3}{4}x + 3 \]

### Step 4: Find the intersection of line AB with the circle
The equation of the circle is:
\[ x^2 + y^2 = 16 \]
Substituting \(y\) from the line equation into the circle equation:
\[ x^2 + \left(-\frac{3}{4}x + 3\right)^2 = 16 \]
Expanding this:
\[ x^2 + \left(\frac{9}{16}x^2 - \frac{9}{2}x + 9\right) = 16 \]
Combining terms:
\[ \left(1 + \frac{9}{16}\right)x^2 - \frac{9}{2}x + 9 - 16 = 0 \]
This simplifies to:
\[ \frac{25}{16}x^2 - \frac{9}{2}x - 7 = 0 \]
Multiplying through by 16 to eliminate the fraction:
\[ 25x^2 - 72x - 112 = 0 \]

### Step 5: Solve the quadratic equation for x
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{72 \pm \sqrt{(-72)^2 - 4 \cdot 25 \cdot (-112)}}{2 \cdot 25} \]
Calculating the discriminant:
\[ x = \frac{72 \pm \sqrt{5184 + 11200}}{50} = \frac{72 \pm \sqrt{16384}}{50} = \frac{72 \pm 128}{50} \]
This gives us two solutions:
\[ x_1 = \frac{200}{50} = 4, \quad x_2 = \frac{-56}{50} = -\frac{28}{25} \]

### Step 6: Find the corresponding y-coordinate for \(x_2\)
Substituting \(x_2 = -\frac{28}{25}\) into the line equation:
\[ y = -\frac{3}{4}\left(-\frac{28}{25}\right) + 3 = \frac{21}{25} + 3 = \frac{21}{25} + \frac{75}{25} = \frac{96}{25} \]
Thus, point P is:
\[ P\left(-\frac{28}{25}, \frac{96}{25}\right) \]

### Step 7: Calculate the area of triangle OAP
Using the formula for the area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting \(O(0,0)\), \(A(4,0)\), and \(P\left(-\frac{28}{25}, \frac{96}{25}\right)\):
\[ \text{Area} = \frac{1}{2} \left| 0(0 - \frac{96}{25}) + 4\left(\frac{96}{25} - 0\right) + \left(-\frac{28}{25}\right)(0 - 0) \right| \]
This simplifies to:
\[ \text{Area} = \frac{1}{2} \left| 4 \cdot \frac{96}{25} \right| = \frac{192}{25} \]

### Final Answer
The area of triangle OAP is:
\[ \frac{192}{25} \text{ square units} \]