It is known that the eccentricity of hyperbola (x ^ 2 / A ^ 2) - (y ^ 2 / 4) = 1 (a > 0) is root 2 and the focus of parabola y ^ 2 = 2px (P > 0) is at the top of hyperbola It is known that the eccentricity of hyperbola (x ^ 2 / A ^ 2) - (y ^ 2 / 4) = 1 (a > 0) is root 2, and the focus of parabola y ^ 2 = 2px (P > 0) is on the apex of hyperbola to solve parabolic equation?

It is known that the eccentricity of hyperbola (x ^ 2 / A ^ 2) - (y ^ 2 / 4) = 1 (a > 0) is root 2 and the focus of parabola y ^ 2 = 2px (P > 0) is at the top of hyperbola It is known that the eccentricity of hyperbola (x ^ 2 / A ^ 2) - (y ^ 2 / 4) = 1 (a > 0) is root 2, and the focus of parabola y ^ 2 = 2px (P > 0) is on the apex of hyperbola to solve parabolic equation?

Eccentricity e = C / a = √ 2
∴c²/a²＝(b²＋a²)/a²＝(4＋a²)/a²＝2
The solution is a = 2
The vertex of hyperbola (2,0) is the focus of parabola
∴p/2＝2
The solution is p = 4
The parabolic equation is Y & # 178; = 8x

If the focus of the parabola y2 = 2px coincides with the right focus of the hyperbola x23 − y2 = 1, then the value of P is ()
A. -4B. 4C. -2D. 2

In ∵ hyperbola x23 − y2 = 1, A2 = 3, B2 = 1 ∵ C = A2 + B2 = 2, the right focus of hyperbola is f (2,0), so the focus of parabola y2 = 2px (p2,0) is f (2,0) ∵ P2 = 2, that is p = 4, so B is selected

If the left focus of the hyperbola x2 / 3 (one third of the square of x) - 16y2 / P2 = 1 (P > 0) is on the Quasilinear of the parabola y2 = 2px, then the eccentricity of the hyperbola is?

Quasilinear equation of parabola y ^ 2 = 2px (P > 0): x = - P / 2
Left focus (- C, 0) of hyperbola, C ^ 2 = 3 + P ^ 2 / 16
3+p^2/16=(p/2)^2
p=4
So C = 2
So the eccentricity of hyperbola: e = C / a = 2 / √ 3 = 2 √ 3 / 3

If the left focus of the square of one-third x minus the square of one-third of P is on the Quasilinear where the square of parabola y equals 2px, then p =?

Is there no positive or negative limit to p?
If P is positive, then p = 4
If P is negative, then p = - 4