We have an ellipse problem. Given points A at coordinates (0,3) and P at coordinates (3, three-halves) lie on ellipse C with equation x squared over a squared plus y squared over b squared equals 1, where a is greater than b and both are positive. We need to find the eccentricity of the ellipse and determine the equation of line l passing through P.
To find the ellipse parameters, we substitute the given points into the ellipse equation. First, using point A at (0,3), we get 9 over b squared equals 1, so b equals 3. Next, using point P at (3, three-halves) with b equals 3, we get 9 over a squared plus one-fourth equals 1, which gives us a squared equals 12, so a equals 2 root 3.
Now we calculate the eccentricity. With a squared equals 12 and b squared equals 9, we find c squared equals a squared minus b squared, which is 3. So c equals root 3. The eccentricity e equals c over a, which is root 3 over 2 root 3, simplifying to one-half. The foci are located at distance c from the center along the major axis.
To find line l, we use the area constraint. The area of triangle ABP with vertices A, P, and B is given by the determinant formula. Substituting the coordinates and simplifying, we get the area equals one-half times the absolute value of three-halves x_B plus 3 y_B minus 9. Since the area equals 9, we have two cases: x_B plus 2 y_B equals 18, or x_B plus 2 y_B equals negative 6.
Solving the constraint x plus 2y equals negative 6 with the ellipse equation, we find two intersection points: B1 at (-3, -3/2) and B2 at (0, -3). The line through P and B1 has equation y equals one-half x. The line through P and B2 has equation y equals three-halves x minus 3. Therefore, the eccentricity is one-half, and the two possible line equations are y equals one-half x and y equals three-halves x minus 3.