Answer:
[tex]\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1[/tex]
Step-by-step explanation:
The equation of the horizontal hyperbola in standard form is:
[tex]\frac{(x-k)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1[/tex]
The position of its center is:
[tex]C(x,y) = \left(\frac{-10 + 70}{2}, -15 \right)[/tex]
[tex]C(x, y) = (30,-15)[/tex]
The values for c and a are respectively:
[tex]a = 70 - 30[/tex]
[tex]a = 40[/tex]
[tex]c = 30 - (-11)[/tex]
[tex]c = 41[/tex]
The remaining variable is computed from the following Pythagorean identity:
[tex]c ^{2} = a^{2} + b^{2}[/tex]
[tex]b = \sqrt{c^{2}-a^{2}}[/tex]
[tex]b = \sqrt{41^{2}-40^{2}}[/tex]
[tex]b = 3[/tex]
Now, the equation of the hyperbola is:
[tex]\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1[/tex]
Answer:
The above answer is correct but the 3 should be a 9
Step-by-step explanation:
Plato
