The type of the conic section using the Discriminant Test and plot the curve using...

Concept Used: The general equation of conic ${\displaystyle a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0}$ has the discriminant ${\displaystyle \mathrm{\Delta }}$ and its given as ${\displaystyle \mathrm{\Delta }=\mid ahghbfgfc\mid =abc+2fgh-a{f}^2-b{g}^2-c{h}^2}$ The general equation of conic ${\displaystyle a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0}$ can also be written in the following way ${\displaystyle A{x}^2+BXY+C{y}^2+Dx+Ey+F=0}$ If ${\displaystyle \mathrm{\Delta }\ne q0,}$ then the equation is nondegenerate and under this case ${\displaystyle var\mathrm{\neg }h\in g\text{If}{B}^2-4AC>0,}$ it represents a hyperbola and a rectangular hyperbola if ${\displaystyle (A+C=0)}$
${\displaystyle var\mathrm{\neg }h\in g\text{If}{B}^2-4AC=0,}$ it represents a parabola. ${\displaystyle var\mathrm{\neg }h\in g\text{If}{B}^2-4AC<0,\text{it represents a circle }(A=C,B=0)}$ or ellipse ${\displaystyle A\ne qC}$ Calculations: The given equation is: ${\displaystyle x^2-2xy+y^2+24x-8=0}$ and the equation is nondegenerate, hence ${\displaystyle \mathrm{\Delta }\ne q0}$. Now let's find the value of ${\displaystyle B^2-4AC}$. From the given equation we have ${\displaystyle A=1,B=-2,C=1,}$ so on substitution we get ${\displaystyle \Rightarrow B^2-4AC={(-2)}^2-4\cdot 1\cdot 1\Rightarrow B^2-4AC=4-4\Rightarrow B^2-4AC=0}$ Hence the given equation represents a Parabola. The plot of the given curve is as below: Conclusion: The given equation represents a Parabola and the plot is shown above.