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