How will u classify the following curves in terms of parabola, ellipse or hyperbola?
i) $\displaystyle 17x^2+12xy+8y^2-46x-28y+33=0$
ii) $\displaystyle x^2-5xy+y^2+8x-20y+15=0$
Hello, varunnayudu!
If you are assigned "rotated" conics, you should have been given more information.How will u classify the following curves as parabola, ellipse or hyperbola?
$\displaystyle 1)\;\;17x^2+12xy+8y^2-46x-28y+33\:=\:0$
$\displaystyle 2)\;\;x^2-5xy+y^2+8x-20y+15\:=\:0$
We know the curve is rotated due to the appearance of the $\displaystyle xy$-term.
The general quadratic has the form: .$\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F \;=\;0$
. . It has a "discriminant": .$\displaystyle \Delta \:=\:B^2-4AC$ . ... look familiar?
. . . . . . . . $\displaystyle \begin{array}{|c|c|}\hline \Delta \:=\:0 & \text{parabola} \\ \hline \Delta \:<\:0 & \text{ellipse} \\ \hline \Delta \:>\:0 & \text{hyperbola}\\ \hline \end{array}$
I have done it this way .Is this right.........
i) $\displaystyle 17x^2 \ + \ 12xy \ + \ 8 y^2 \ - \ 46x \ - \ 28y \ + 33 \ = \ 0$
$\displaystyle a \ = \ 17 ; \ b \ = \ 8 ; h \ = \ 6;$
$\displaystyle ab-h^2 \ = \ 136 - 36 = \ 100 $
since $\displaystyle ab-h^2 > 0 $ therefore the eq is an hyperbola.
ii) $\displaystyle x^2 - 5xy + y^2 + 8x - 20y +15 = 0$
$\displaystyle a=1 ; \ b = 1; \ h = - \frac{5}{2}$
$\displaystyle \therefore \ ab-h^2 = 1 - \frac{25}{4} = \frac{4 - 25}{4} = - \frac{21}{4} \ = -5.25 \ < \ 0$
$\displaystyle \therefore since \ ab-h^2 < 0 $ it is an ellipse.