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For a National Board Exam Review:

What conic section is ${ 2x^2 -8xy+4x+12 }$ ?

Answer is Hyperbola.

But I can't seem to translate it properly to the standard form of a hyperbola.. What am I doing wrong?

$${ 2x^2 - 8xy + 4x = 12 }$$ $${ 2x^2 - 4x(2y-1) = 12 }$$

Divide by -4x

$${ - \frac{1}{2}x - (2y-1) = -3x }$$

I dont know how its going to be a hyperbola with no second degree term...

james
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1 Answers1

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Notice, we have $$x^2-8xy+4x+12$$ Compare the equation with the general form of second degree equation: $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ we get $$a=1, h=-4, b=0, g=2, f=0, c=12 $$ Now, checking if the given curve is a conic section or a pair of straight lines by using discriminant $\Delta$ as follows $$\color{blue}{\Delta=abc+2fgh-af^2-bg^2-ch^2}$$ $$=(1)(0)(12)+2(0)(2)(-4)-(1)(0)^2-(0)(2)^2-(12)(-4)^2$$ $$=-192\neq 0$$

Since, $\Delta\neq 0$ hence, the given curve $x^2-8xy+4x+12$ represents a $\color{red}{\text{conic section}}$.

Now, checking the nature of conic section by using discriminant $$\color{blue}{D=h^2-ab}$$ $$=(-4)^2-(1)(0)=16>0$$ Since, $D=16>0$ Hence, $\color{red}{x^2-8xy+4x+12}$ represents a $\color{red}{\text{Hyperbola}}$

Harish Chandra Rajpoot
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  • Is it possible to get this equation in standard form ? – mathemather Aug 20 '15 at 08:08
  • Yes, but you will have to compare the given curve with the standard form of hyperbola as follows $$x^2-8xy+4x+12=\frac{(x-x_1)^2}{a^2}-\frac{(y-y_1)^2}{b^2}-1$$ Which seems difficult to find out unknowns. So use discriminant to check the nature of hyperbola as explained in the answer. – Harish Chandra Rajpoot Aug 20 '15 at 08:17