Question

Find an equation for the hyperbola that satisfies the given conditions.
Asymptotes: y = ±x,
hyperbola passes through (5, 3)

Answer 1

Answer:

$\frac{x^{2}}{16} - \frac{y^{2}}{16} = 1$

Step-by-step explanation:

The slope of the one of the asymptotes is:

$\frac{b}{a} = 1$

$a = b$

The form of the asymptotes indicates that the hyperbola is centered at origin. Therefore, the equation of the conic is simplified into this form:

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

$x^{2}-y^{2} = a^{2}$

$a = \sqrt{5^{2}-3^{2}}$

$a = 4$

The equation of the hyperbola is:

$\frac{x^{2}}{16} - \frac{y^{2}}{16} = 1$

Answer 2

The equation of hyperbola that satisfies the given conditions is $\dfrac{x^2}{16}-\dfrac{y^2}{16} =1$ and this can be determined by using the given data.

Given :

• Asymptotes: y = ±x,

• Hyperbola passes through (5, 3)

The generalized equation of Hyperbola is given below:

$\rm \dfrac{(x-h)^2}{a^2}- \dfrac{(y-k)^2}{b^2}=1$   --- (1)

So, according to the data, the asymptotes: y = ±x. Therefore, the slope is:

$\rm \dfrac{b}{a}=1$

a = b

The value of h = 0 and k = 0 in the equation (1).

Now, in order to determine the value of 'a' use the below calculation.

$\dfrac{x^2}{a^2}-\dfrac{y^2}{a^2} =1$

$x^2-y^2=a^2$

Now, substitute the value of y and x in the above equation.

$5^2-3^2=a^2$

25 - 9 = $a^2$

a = 4

So, the value of 'b' is also equal to 4.

Now, substitute the values of the known terms in the equation (1).

$\dfrac{x^2}{16}-\dfrac{y^2}{16} =1$

For more information, refer to the link given below:

https://brainly.com/question/12919612