Question

The center of a hyperbola is (−2,4) , and one vertex is (−2,7) . The slope of one of the asymptotes is 12 .

What is the equation of the hyperbola in standard form?

Answer 1

Answer:

The equation of the hyperbola in standard form is $\frac{(x - 4)^2 }{ 9} - \frac{(y +2)^2}{2.25} = 1$

Step-by-step explanation:

Given:

Centre of the hyperbola=(−2,4)

one vertex of the hyperbola= (−2,7) .

slope of the asymptote = 12

To Find:

The equation of the hyperbola in standard form=?

Solution:

W know that the  standard form of hyper bola is

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

where

(h,k) is the centre

(x,y) is the vertex of the parabola

a is the length between the centre and the vertices of the hyperbola

b is  the distance perpendicular to the transverse axis from the vertex to the asymptotic line

Now the length of a is given by

a=|k-y|

a=|4-7|

a=|-3|

a=3

Also we know that,

Slope =$\frac{a}{b}$= 2

=>$\frac{3}{b}=2$

=>$\frac{3}{2}=b$

=>b=1.5

Now substituting the known values in equation(1)

$\frac{(x - 4)^2 }{ 3^2} - \frac{(y - (-2)^2}{  1.5^2} = 1$

$\frac{(x - 4)^2 }{ 9} - \frac{(y +2)^2}{2.25} = 1$

Answer 2

Answer:

$\frac{(x+2)^{2}}{1296} - \frac{(y-4)^{2}}{9} = 1$

Step-by-step explanation:

The standard form of a hyperbola centered at a point distinct from origin has the following form:

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

The distance between the center and the vertex is:

$b = \sqrt{[(-2)-(-2)]^{2}+(7-4)^{2}}$

$b = 3$

The value of the other semiaxis is:

$\frac{a}{b} = 12$

$a = 12\cdot b$

$a = 36$

The standard equation of the hyperbola is:

$\frac{(x+2)^{2}}{1296} - \frac{(y-4)^{2}}{9} = 1$