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How do you find the center of the hyperbola, its focal length, and its eccentricity if a hyperbola has a vertical transverse axis of length $8$ and asymptotes of $y = \dfrac{7}{2}x - 3$ and $y = - \dfrac{7}{2}x - 1?$

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint:With the help of two given equations, will find the center of the hyperbola and then will find the slope and eccentricity. Also, will use the substitution method to find the unknown terms.

Complete step by step solution:
We know that the hyperbola has a vertical transverse axis equal to $( - 1)$ at the second member with the center in $C({x_c},{y_c})$ which is not the origin.
Asymptotes do not pass from the origin; the equation can be expressed as:
$\dfrac{{{{(x - {x_c})}^2}}}{{{a^2}}} - \dfrac{{{{(y - {y_c})}^2}}}{{{b^2}}} = ( - 1)$ ….. (I)
Find the center which is at the point and it is on the two asymptotes. To get the center of the points, solve the below two equations.
$y = \dfrac{7}{2}x - 3$ …(A)
and $y = - \dfrac{7}{2}x - 1$… (B)
Add equations (A) and (B)
$ \Rightarrow 2y = - 3 - 1$ (Since like terms with the same value and the opposite sign cancel each other)
Now, when there are two negative numbers, you have to add the numbers and give minus sign to the resultant value.
$ \Rightarrow 2y = - 4$
When any term multiplicative on one side is moved to opposite side, then it goes to the
denominator.
$ \Rightarrow y = \dfrac{{ - 4}}{2}$
Common multiples from the numerator and the denominator cancel each other.
$ \Rightarrow y = ( - 2)$
Place the value of the above equation in the equation (A)
$( - 2) = \dfrac{7}{2}x - 3$
When any term is moved from one side to another, the sign of the term also changes. Negative terms become positive and vice-versa.
$ \Rightarrow - 2 + 3 = \dfrac{7}{2}x$
Simplify the above equation.
$ \Rightarrow 1 = \dfrac{7}{2}x$
Do cross multiplication, where the numerator of one side is multiplied with the denominator of the opposite side.
$ \Rightarrow 2 = 7x$
Make unknown the subject.
$ \Rightarrow x = \dfrac{2}{7}$
Therefore, the center is $\left( {\dfrac{2}{7},0} \right)$
The slope of asymptotes can be expressed as: $m = \pm \dfrac{b}{a}$ …. (c)
Also, the vertical axis given is $8$
$
\therefore 2b = 8 \\
\Rightarrow b = 4 \\
$
Now, place values in equation (C)
$ \Rightarrow \dfrac{4}{a} = \dfrac{7}{2}$
Do cross multiplication
$ \Rightarrow 7a = 8$
Make “a” as the subject
$ \Rightarrow a = \dfrac{8}{7}$
Place all the values in equation (I)
$\dfrac{{{{\left( {x - \dfrac{2}{7}} \right)}^2}}}{{{{\left( {\dfrac{8}{7}} \right)}^2}}} - \dfrac{{{{(y + 2)}^2}}}{{{4^2}}} = ( - 1)$
Simplify the above equation:
$\dfrac{{{{\left( {x - \dfrac{2}{7}} \right)}^2}}}{{\dfrac{{64}}{{49}}}} - \dfrac{{{{(y + 2)}^2}}}{{16}} = ( - 1)$
Also, the focal axis is named as $2c$ in such a way that ${c^2} = {a^2} + {b^2}$
Place the values and simplify
$\therefore c = \sqrt {\dfrac{{64}}{{49}} + 16} $
Simplify
$
\therefore c = \sqrt {\dfrac{{64 + 784}}{{49}}} \\
\therefore c = \sqrt {\dfrac{{848}}{{49}}} \\
\therefore c = \sqrt {\dfrac{{16 \times 53}}{{49}}} \\
\therefore c = \dfrac{4}{7}\sqrt {53} \\
$
Therefore, the eccentricity will be:
$e = \dfrac{c}{b}$
Place values and simplify
$e = \dfrac{{\dfrac{4}{7}\sqrt {53} }}{4}$
Common factors from the numerator and the denominator cancel each other.
$e = \dfrac{{\sqrt {53} }}{7}$
This is the required solution.

Note: Always be careful about the sign convention while simplification of the equations. When you move any term from one side to the other, the sign of the term is changed. Positive terms are changed to negative and vice-versa.