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# Hyperbola Calculator

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Hyperbola is a curve with a set of points in a plan such that the distance from those points to the two fixed points are constant. Here, the two fixed lines are the foci.

If the coordinate points and value of a and b are given, here a and b are the points from center of hyperbola,
then the hyperbola calculator helps calculate the focii, eccentricity and the asymptotes.

## Steps

Step 1 : Check whether the given problem is in the form,
$\frac{(x - x_{0})^{2}}{a^{2}}-\frac{(y - y_{0})^{2}}{b^{2}}$ = 1
where, a and b are the points from center of hyperbola and find the center points.

Step 2 : To get the desired parameter, substitute the values in the following equations given below.

Use the following equations of hyperbola,
Focus F of X coordinate = $x_{0}+\sqrt{(a^2+b^2)}$

Focus F of Y coordinate = $y_{0}$
Focus F' of X coordinate = $x_{0}-\sqrt{(a^2+b^2)}$

Focus F' of Y coordinate =$y_0$
Asymptotes for L'H :

Asymptotes for H'L : y = -$\frac{b}{a}$$x + y_{0} + \frac{b}{a}$$x_0$

Eccentricity e = $\frac{\sqrt{(a^2+b^2)}}{a}$

## Problems

Below are the problems based on hyperbola.

### Solved Examples

Question 1:
Find the center, foci, asymptotes and eccentricity of the given equation,
$\frac{(x - 3)^{2}}{25} - \frac{(y + 1)^{2}}{49}$ = 1

Solution:
Step 1: Given equation, $\frac{(x - 3)^{2}}{25}-\frac{(y + 1)^{2}}{49}$ = 1
The center points are (3, -1) and a = 5 , b = 7.

Step 2 :
Focus F of X coordinate = $x_{0} + \sqrt{(a^2+b^2)}$

= 3 + $\sqrt{(5^2 + 7^2)}$
= 3 + 8.602
= 11.602

Focus F of Y coordinate = -1

Focus F' of X coordinate = $x_{0} - \sqrt{(a^2+b^2)}$

= 3 - $\sqrt{(5^2 + 7^2)}$
= 3 - 8.602
= -5.602

Focus F' of Y coordinate = -1

Asymptotes for L'H : y = $\frac{b}{a}$$x + y_{0} - \frac{b}{a}$$x_0$

= $\frac{7}{5}$$x - 1 - \frac{7}{5}$$ \times$ 3

= 1.4x - 5.200

Asymptotes for H'L : y = -$\frac{b}{a}$$x + y_{0} + \frac{b}{a}$$x_0$

= -$\frac{7}{5}$$x + 1 - \frac{7}{5}$$ \times$ 3

= -1.4x + 3.200

Eccentricity e = $\frac{\sqrt{(a^2+b^2)}}{a}$

= $\frac{\sqrt{(5^2+7^2)}}{5}$

= $\frac{\sqrt{74}}{5}$

= 1.720

Question 2:
Find the center, foci, asymptotes and eccentricity of the given equation,
$\frac{(x + 3)^{2}}{16}-\frac{(y - 2)^{2}}{9}$ = 1

Solution:
Step 1: Given equation, $\frac{(x + 3)^{2}}{16}-\frac{(y - 2)^{2}}{9}$ = 1
The center points are (-3, 2) and a = 4 , b = 3.

Step 2 :
Focus F of X coordinate = $x_{0} + \sqrt{(a^2+b^2)}$

= -3 + $\sqrt{(4^2 + 3^2)}$
= -3 + 5
= 2

Focus F of Y coordinate = 2

Focus F' of X coordinate = $x_{0} - \sqrt{(a^2+b^2)}$

= -3 - $\sqrt{(4^2 + 3^2)}$
= -3 - 5
= -8

Focus F' of Y coordinate = 2

Asymptotes for L'H : y= $\frac{b}{a}$$x + y_{0} - \frac{b}{a}$$x_0$

= $\frac{3}{4}$$x + 2 - \frac{3}{4}$$ \times$ (-3)

= 0.75x + 4.25

Asymptotes for H'L : y = -$\frac{b}{a}$$x + y_{0} + \frac{b}{a}$$x_0$

= -$\frac{3}{4}$$x + 2 + \frac{3}{4}$$ \times$ (-3)

= -0.75x - 0.25

Eccentricity e = $\frac{\sqrt{(a^2+b^2)}}{a}$

= $\frac{\sqrt{(4^2+3^2)}}{4}$

= $\frac{\sqrt{25}}{4}$

= 1.25