Let the asymptote of hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0) be tangent to the parabola y = x2 + 1, then the eccentricity of the hyperbola is equal to () A. 3B. 2C. 5D. 6

Let the asymptote of hyperbola x2a2 − y2b2 = 1 (a > 0, b > 0) be tangent to the parabola y = x2 + 1, then the eccentricity of the hyperbola is equal to () A. 3B. 2C. 5D. 6

The asymptote equation of hyperbola x2a2 − y2b2 = 1 (a ＞ 0, B ＞ 0) is y = BXA, which is substituted into the parabolic equation to sort out ax2-bx + a = 0. Because the asymptote is tangent to the parabola, b2-4a2 = 0, that is, C2 = 5A2 ⇔ e = 5, so C

It is known that a focus of hyperbola C: x ^ 2 / A ^ + y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) is F2 (2,0), and B = root 3a. (1) let the equation (2) of hyperbola C be solved

First of all, you copy the wrong question
The focus is that F2 (2,0) obtains C = 2, because it is hyperbola C2 = A2 + B2, that is, 4 = A2 + 3a2, and the solution obtains a = 1, B = radical 3

What is the equation of a parabola with the right vertex of the hyperbola as the vertex and the left focus as the focus?
The major axis of the hyperbola is on the X axis
a^2=16,b^2=9..

a^2=16,b^2=9
Right vertex (4,0), left focus (- 5,0) of hyperbola
Parabola opening left
Let the parabolic equation be
y^2=-2p(x-4) (p>0)
p/2=4-(-5)=9
2p=36
The parabolic equation is
y^2=-36(x-4)

Given a point on the hyperbola X & sup2 / 225-y & sup2 / 64 = 1, its abscissa is equal to 15, try to find the distance from the point to two focal points

If x = 15 is substituted into y = 0, then the point is (15,0),
C & sup2; = A & sup2; + B & sup2; = 225 + 64 = 289, C = 17, the focus coordinates are (- 17,0), (17,0)
L1=15-(-17)=32,L2=17-15=2