Question

Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and (10,0) The equation in standard form?

Answer 1

Answer:

$\frac{x^{2}}{36} - \frac{y^{2}}{64}=1$

Step-by-step explanation:

Given an hyperbola with the following conditions:

• Foci at (-10,0) and (10,0)
• x-intercept +/- 6;

The following holds:

• The center is midway between the foci, so the center must be at (h, k) = (0, 0).

• The foci are 10 units to either side of the center, so c = 10 and $c^2 = 100$
• The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and $a^2 = 36.$

$Then, a^2+b^2=c^2\\b^2=100-36=64$

Therefore, substituting $a^2 = 36.$ and $b^2=64$ into the standard form

$\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1$