Question

Trace the following central conics.
$4{{x}^{2}}+27xy+35{{y}^{2}}-14x-31y-6=0$.

Answer 1

We will compare the given equation to the general equation of second degree. The general equation of second degree is as follows, $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$. Then we will look at the equation $\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}$. We will also check the relation between ${{h}^{2}}$ and $ab$. With the help of these three equations, and their relations, we will be able to determine the conics section represented by the given equation.

Complete step by step answer:
The given equation is $4{{x}^{2}}+27xy+35{{y}^{2}}-14x-31y-6=0$. The general equation of second degree is $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$. Comparing these two equations, we get the following values, $a=4$, $b=35$, $c=-6$, $h=\dfrac{27}{2}$, $f=\dfrac{-31}{2}$ and $g=-7$.
We have $\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}$. Now, we will substitute the values we have in $\Delta$. We get the following equation,
$\Delta =(4\times 35\times -6)+2\left( \dfrac{-31}{2}\times -7\times \dfrac{27}{2} \right)-4\times {{\left( \dfrac{-31}{2} \right)}^{2}}-35\times {{\left( -7 \right)}^{2}}+6\times {{\left( \dfrac{27}{2} \right)}^{2}}$
Simplifying this equation, we get
$\begin{aligned} & \Delta =-840+2929.5-961-1715+1093.5 \\\ & =507 \end{aligned}$
Next, we calculate ${{h}^{2}}$ and $ab$. So we have, ${{h}^{2}}={{\left( \dfrac{27}{2} \right)}^{2}}=\dfrac{729}{4}=182.25$ and $ab=4\times 35=140$.
Now, as $\Delta \ne 0$ and ${{h}^{2}}>ab$, we conclude that the given equation is a hyperbola. The coordinates of the centre of the hyperbola are $(0,0)$. The coordinates of the vertices are $(a,0)$ and $(-a,0)$, that is, $(4,0)$ and $(-4,0)$. The eccentricity is given by the following formula, $e=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}$ . Substituting the values of $a$ and $b$, we get the following,
$e=\sqrt{\dfrac{{{(4)}^{2}}+{{(35)}^{2}}}{{{(4)}^{2}}}}=\sqrt{\dfrac{16+1225}{16}}=\sqrt{\dfrac{1241}{16}}=\sqrt{77.5625}=8.807$.
The coordinates of the foci are $(\pm ae,0)$, that is $(\pm 35.228,0)$. The graph that we get by plotting the given equation is as follows,

Note:
The values of $\Delta$, ${{h}^{2}}$ and $ab$ help us in determining which conics section is being represented by the given equation. If $\Delta =0$, they are a pair of straight lines. If $\Delta \ne 0,a=b$ and $h=0$, then it is a circle. If $\Delta \ne 0$ and ${{h}^{2}}=ab$, then it is a parabola. If $\Delta \ne 0$ and ${{h}^{2}}ab$, then it is a hyperbola.