As shown in the figure, the left and right fixed points of ellipse C1: x ^ 2 / 4 + y ^ 2 / 3 = 1 are a, B and P respectively, which are a point on the right branch (above the X axis) of hyperbola C2: x ^ 2 / 4-y ^ 2 / 3 = 1 Connecting AP is called C1 to C, connecting Pb and extending the intersection C1 to D, and the areas of △ ACD and △ PCD are equal. Calculate the slope of the straight line PD and the inclination angle of the straight line CD

As shown in the figure, the left and right fixed points of ellipse C1: x ^ 2 / 4 + y ^ 2 / 3 = 1 are a, B and P respectively, which are a point on the right branch (above the X axis) of hyperbola C2: x ^ 2 / 4-y ^ 2 / 3 = 1 Connecting AP is called C1 to C, connecting Pb and extending the intersection C1 to D, and the areas of △ ACD and △ PCD are equal. Calculate the slope of the straight line PD and the inclination angle of the straight line CD

The areas of △ ACD and △ PCD are equal
Two triangles of equal height
∴AC=PC
Let the coordinates of P and C be (x0, Y0), (x1, Y1) respectively
A. The coordinates of B are (- 2,0) (2,0)
(-2+x0)/2=x1
(0+y0)/2=y1
Substituting C1: x ^ 2 / 4 + y ^ 2 / 3 = 1
3(x0-2)^2+4y0^2=48
P (x0, Y0) is still there
C2: on x ^ 2 / 4-y ^ 2 / 3 = 1
∴x0^2/4-y0^2/3=1
2 simultaneous
Ψ x0 = 4 or - 2 (rounding off)
y0=3
The coordinates of point P are (4,3)
C(1,3/2)
PD slope = (3-0) / (4-2) = 3 / 2
Pb analytic formula y = 3 / 2 (X-2)
Combined with x ^ 2 / 4 + y ^ 2 / 3 = 1
x^2-3x+2=0
(x-1)(x-2)=0
The abscissa of D is 1
The CD is x = 1
The inclination angle is 90 degrees