Find the Properties x^2-4y^2=4

Problem

x2−4*y2=4

Solution

1. Rewrite the equation in standard form by dividing both sides by 4

(x2)/4−(4*y2)/4=4/4

(x2)/4−(y2)/1=1

2. Identify the type of conic section. Since the equation is in the form (x2)/(a2)−(y2)/(b2)=1 it is a horizontal hyperbola.

3. Determine the values of a and b

a2=4⇒a=2

b2=1⇒b=1

4. Calculate the value of c using the relation c2=a2+b2

c2=4+1=5

c=√(,5)

5. Find the center, vertices, and foci. The center is at the origin (0,0) The vertices are at (±a,0) and the foci are at (±c,0)

Center:(0,0)

Vertices:(2,0),(−2,0)

Foci:(√(,5),0),(−√(,5),0)

6. Determine the asymptotes using the formula y=±b/a*x

y=±1/2*x

7. Calculate the eccentricity using e=c/a

e=√(,5)/2

Final Answer

Properties of *x2−4*y2=4:Center *(0,0),Vertices *(±2,0),Foci *(±√(,5),0),Asymptotes *y=±1/2*x

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