Question

The eccentricity of the hyperbola $9{{x}^{2}}-16{{y}^{2}}-18x-64y-199=0$ is
A. $\dfrac{16}{9}$
B. $\dfrac{5}{4}$
C. $\dfrac{25}{16}$
D. 0

Answer 1

Hint : We first explain the mathematical aspect of the conic curve hyperbola. We find the general formula and explain different components of the curve. Then we equate it with the given curve to find the eccentricity of $9{{x}^{2}}-16{{y}^{2}}-18x-64y-199=0$.

Complete step-by-step answer :
The general equation hyperbola is $\dfrac{{{\left( x-\alpha \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-\beta \right)}^{2}}}{{{b}^{2}}}=1$.
For the general equation $\left( \alpha ,\beta \right)$ is the centre. The vertices are $\left( \alpha \pm a,\beta \right)$. The coordinates of the foci are $\left( \alpha \pm ae,\beta \right)$. Here $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ is the eccentricity.
We now convert $9{{x}^{2}}-16{{y}^{2}}-18x-64y-199=0$ to the general form.
$\begin{aligned} & 9{{x}^{2}}-16{{y}^{2}}-18x-64y-199=0 \\\ & \Rightarrow {{\left( 3x-3 \right)}^{2}}-{{\left( 4y+8 \right)}^{2}}=144={{12}^{2}} \\\ & \Rightarrow \dfrac{{{\left( 3x-3 \right)}^{2}}}{{{12}^{2}}}-\dfrac{{{\left( 4y+8 \right)}^{2}}}{{{12}^{2}}}=1 \\\ & \Rightarrow \dfrac{{{\left( x-1 \right)}^{2}}}{16}-\dfrac{{{\left( y+2 \right)}^{2}}}{9}=1 \\\ \end{aligned}$
Therefore, the eccentricity is
$e=\sqrt{1+\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}$.
So, the correct answer is “Option B”.

Note : A hyperbola is the mathematical shape in the form of a smooth curve formed by the intersection of two circular cones. The properties of hyperbola allow it to play an important role in the real world where designs and predictions of phenomena are heavily influenced by it.
The hyperbola has an important mathematical equation associated with it - the inverse relation. When an increase in one trait leads to a decrease in another or vice versa which helps in explaining the relationship between the pressure and volume of a gas.