Question

How do you identify the conic $4{x^2} + 8{y^2} - 8x - 24 = 4$ is, if any and if the equation does represent a conic, state its vertex or center?

Answer 1

Hint:
To check whether the given equation is that of a conic or not we compare it with the general form of the conic given by, $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$. An equation of a conic is a second degree equation, i.e. both $a$ and $b$ cannot be $0$ at the same time. Once we establish that given equation is that of a conic we will use the discriminant method to check for which conic is represented by the given equation using the following table:

Condition	Type of conic
$\Delta = 0$	Pair of straight lines
$\Delta \ne 0$, and$D < 0$, and$a = b$, and$h = 0$	Circle
$\Delta \ne 0$, and$D < 0$, and$a \ne b$	Ellipse
$\Delta \ne 0$, and$D = 0$	Parabola
$\Delta \ne 0$, and$D > 0$	Hyperbola

where, determinant $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$
discriminant $D = {h^2} - ab$

Complete step by step solution:
The given equation is $4{x^2} + 8{y^2} - 8x - 24 = 4$
We shift all the terms to the LHS,
$
  4{x^2} + 8{y^2} - 8x - 24 - 4 = 0 \\
   \Rightarrow 4{x^2} + 8{y^2} - 8x - 28 = 0 \\
 $
On comparing with the general form of the conic given by,
$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
We see that,
$a = 4,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} b = 8,{\kern 1pt} {\kern 1pt} {\kern 1pt} c = - 28,{\kern 1pt} {\kern 1pt} {\kern 1pt} g = - 4$ and $h = f = 0$
Since, the equation is a second-degree equation, and both $a$ and $b$ are not $0$, we can say that this equation represents a conic.
Now we have to check for which conic is represented by this equation. For this we will use the discriminant method and use the above given table.
First we calculate $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$,
$
  \Delta = (4 \times 8 \times - 28) + (2 \times 0 \times - 4 \times 0) - (4 \times 0 \times 0) - (8 \times - 4 \times - 4) - ( - 28 \times 0 \times 0) \\
   \Rightarrow \Delta = ( - 896) + 0 - 0 - 128 - 0 \\
   \Rightarrow \Delta = - 896 - 128 = - 1024 \\
 $
Thus, $\Delta \ne 0$.
Now we calculate $D = {h^2} - ab$,
$
  D = {0^2} - (4 \times 8) \\
   \Rightarrow D = - 32 \\
 $
Thus, $D < 0$.
So, from the table we see that the given equation can be that of a circle or an ellipse.
We further compare $a$ and $b$. Since, $a = 4$ and $b = 8$, we have $a \ne b$.
The given equation satisfies the following conditions:
$\Delta \ne 0$, $D < 0$, and $a \ne b$.
Thus, from the table we see that the given equation satisfies all the conditions of that of an ellipse.
To get the vertices and center we rewrite the equation in the standard form of the ellipse, i.e. $\dfrac{{{{(x - \alpha )}^2}}}{{{a^2}}} + \dfrac{{{{(y - \beta )}^2}}}{{{b^2}}} = 1$.
\[
  4{x^2} + 8{y^2} - 8x - 24 = 4 \\
   \Rightarrow 4{x^2} + 8{y^2} - 8x - 24 - 4 = 0 \\
   \Rightarrow 4{x^2} + 8{y^2} - 8x - 28 = 0 \\
   \Rightarrow 4({x^2} + 2{y^2} - 2x - 7) = 0 \\
   \Rightarrow {x^2} + 2{y^2} - 2x - 7 = 0 \\
   \Rightarrow {x^2} + 2{y^2} - 2x - 7 + 1 - 1 = 0 \\
   \Rightarrow ({x^2} - 2x + 1) + (2{y^2}) - 7 - 1 = 0 \\
   \Rightarrow {(x - 1)^2} + (2{y^2}) - 8 = 0 \\
   \Rightarrow {(x - 1)^2} + (2{y^2}) = 8 \\
   \Rightarrow \dfrac{{{{(x - 1)}^2}}}{8} + \dfrac{{{y^2}}}{4} = 1 \\
   \Rightarrow \dfrac{{{{(x - 1)}^2}}}{8} + \dfrac{{{{(y - 0)}^2}}}{4} = 1 \\
 \]
On comparing with standard form of the ellipse $\dfrac{{{{(x - \alpha )}^2}}}{{{a^2}}} + \dfrac{{{{(y - \beta )}^2}}}{{{b^2}}} = 1$, we get,
$\alpha = 1,{\kern 1pt} {\kern 1pt} {\kern 1pt} \beta = 0$
${a^2} = 8 \Rightarrow a = 2\sqrt 2 $
${b^2} = 4 \Rightarrow b = 2$
Center of the ellipse: $(\alpha ,\beta ) = (1,0)$
Vertices:
\[
  (\alpha + a,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \beta ){\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} (\alpha - a,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \beta ){\kern 1pt} {\kern 1pt} {\kern 1pt}, (\alpha, {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \beta + b){\kern 1pt} {\kern 1pt} {\kern 1pt}, {\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} (\alpha, {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \beta - b) \\
   = {\kern 1pt} {\kern 1pt} {\kern 1pt} (1 + 2\sqrt 2 ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 0){\kern 1pt} {\kern 1pt} {\kern 1pt}, {\kern 1pt} {\kern 1pt} {\kern 1pt} (1 - 2\sqrt 2 ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 0){\kern 1pt} {\kern 1pt} {\kern 1pt} ,(1,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 2){\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} (1,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 2) \\
 \]

Note:
We can also use a short method to identify the type of conic once we establish that it is an equation of a conic and $\Delta \ne 0$ . In the general form of equation,
1) if $a = b \ne 0$, it is an equation of a circle.
2) if $a \ne b$ (both $a$ and $b$ not equal to $0$) and both of them are positive, it is an equation of an ellipse.
3) if $a \ne b$ (both $a$ and $b$ not equal to $0$) and exactly one of them is negative, it is an equation of a hyperbola.
4) if $a = 0$ or $b = 0$ (both $a$ and $b$ not equal to $0$ at the same time), it is an equation of a parabola.