Hurry im timed which is the equation of a hyperbola centered at the origin with focus (0 4) and vertex (0 sqrt12)

Centered at the origin: C=(0,0)=(h,k)→h=0, k=0
Focus: F=(0, 4)
Vertex: V=(0, sqrt 12)

The focus and the vertex are on a vertical line: x=0, then the equation has the form:
y^2/a^2-x^2/b^2=1

Distance between the Center and the Vertex: a=sqrt(12)
a^2=[sqrt(12)]^2→a^2=12

Distance between the Center and the Focus: c=4
c^2=(4)^2→c^2=16

c^2=a^2+b^2
Replacing c^2=16 and a^2=12
16=12+b^2
Solving for b^2: 
16-12=12+b^2-12
4=b^2
b^2=4

Replacing a^2=12 and b^2=4 in the equation of the hyperbola:
y^2/a^2-x^2/b^2=1
y^2/12-x^2/4=1

The equation of the hyperbola is y^2/12-x^2/4=1

Fortnite is good imjaystation

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12