Answer to Question #155292 in Calculus for sakshi

Question

trace the conic x2+2xy+y2-3x+y-2=-0

Answer 1

General Cartesian form of the equation is

$Ax^2 +Bxy+Cy^2 +Dx+Ey+F=0;$

If discriminant is equal to zero, the equation represents a parabola.

$B^2 - 4AC = 2^2 - 4\times1\times1 = 0.$

Therefore it is a parabola.

$x^2+2xy+y^2-3x+y-2=0$

is a parabola with vertex at $(h,k)=(-\frac{3}{4},-\frac{1}{2})$

and focal length  $\left|p\right|=\frac{3}{4}$

$4\cdot (\frac{3}{4})(y-(-\frac{3}{4}))=(x-(-\frac{1}{2}))^2$ is the standand equation of the parabola

Focus : $(0,-\frac{1}{2})$

directrix: $x = - \frac{3}{2}$

vertex: $(-\frac{3}{4},-\frac{1}{2})$

Axis: horizontal

Standard Form:

$4\cdot (\frac{3}{4})(y-(-\frac{3}{4}))=(x-(-\frac{1}{2}))^2$

Below is the graph of conic