Question

Find the equation of the hyperbola with center at the origin, length of conjugate axis 10 and one of the foci $\left( { - 7,0} \right)$.

Answer 1

Hint: Since the hyperbola is origin centered and one of its foci is at $\left( { - 7,0} \right)$, this is a horizontal hyperbola. The length of the conjugate axis of hyperbola is $2b$. Compare its value with the given value and then apply formula ${b^2} = {a^2}\left( {{e^2} - 1} \right)$ to determine the values of $a$ and $b$. Put these values in the general equation of hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$.

Complete step-by-step answer:
According to the question, the center of the hyperbola is at the origin with one of its foci at $\left( { - 7,0} \right)$. This suggests that it is a horizontal hyperbola.
We know that the general equation of a horizontal hyperbola centered at origin is:
$ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }}.....{\text{(1)}}$
So to complete this equation, we need to determine the values of $a$ and $b$.
The length of the conjugate axis is given as 10 in the question. And we know that in hyperbola, the length of the conjugate axis is $2b$. So we have:
$
   \Rightarrow 2b = 10 \\
   \Rightarrow b = 5{\text{ }}.....{\text{(2)}}
 $
Further, we also know that the coordinates of foci of a horizontal, origin centered hyperbola are $\left( { \pm ae,0} \right)$, where $e$ is the eccentricity of the hyperbola.
One of the foci is given in the question as $\left( { - 7,0} \right)$. This means it is representing the focus on the negative x-axis i.e. $\left( { - ae,0} \right)$. From this, we have:
$
   \Rightarrow - ae = - 7 \\
   \Rightarrow {a^2}{e^2} = 49{\text{ }}.....{\text{(3)}}
 $
A relation between $a,{\text{ }}b$ and $e$ in hyperbola is given as:
$ \Rightarrow {b^2} = {a^2}\left( {{e^2} - 1} \right)$
On further simplification, this will give us:
$ \Rightarrow {b^2} = {a^2}{e^2} - {a^2}$
Putting the values \[b\] and ${a^2}{e^2}$ from equation (2) and (3) respectively, we’ll get:
$
   \Rightarrow 25 = 49 - {a^2} \\
   \Rightarrow {a^2} = 49 - 25 = 24{\text{ }}.....{\text{(4)}}
 $
Thus from equation (2), we have ${b^2} = 25$ and from equation (4), ${a^2} = 24$. Putting these values in equation (1), we’ll get the equation of hyperbola:
$ \Rightarrow \dfrac{{{x^2}}}{{24}} - \dfrac{{{y^2}}}{{25}} = 1$

Thus the required equation of the given hyperbola is $\dfrac{{{x^2}}}{{24}} - \dfrac{{{y^2}}}{{25}} = 1$.

Note: The range of values of eccentricity varies with different conic sections:
For ellipse, the value of eccentricity is less than 1 i.e. $e < 1$.
For parabola, its value is exactly 1 i.e. $e = 1$.
And for hyperbola, the value of eccentricity is greater than 1 i.e. $e > 1$.