Question 308992
This is very helpful to remember:



For the general conic {{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}} 


If {{{B^2-4AC<0}}}, then the given conic above is an ellipse
Furthermore, if {{{A=C}}}, and {{{B=0}}}, then the conic is also a circle



If {{{B^2-4AC=0}}}, then the given conic above is a parabola



If {{{B^2-4AC>0}}}, then the given conic above is a hyperbola



First, let's subtract 3 from both sides to get {{{4x^2 + ky^2 - 8x + 17y -3=0}}}



So if we wanted to force {{{4x^2 + ky^2 - 8x + 17y -3=0}}} to be a circle, then we must make sure that {{{B^2-4AC<0}}}, {{{A=C}}}, and {{{B=0}}}. In this case, {{{A=4}}}, {{{B=0}}}, {{{C=k}}}, {{{D=-8}}}, {{{E=17}}}, and {{{F=-3}}}. Plug these values in to get {{{0^2-4(4)(k)<0}}} and simplify to get {{{-16k<0}}}. Solve for 'k' to get {{{k>0}}}. So 'k' must be positive.


Also, because we want {{{A=C}}} and {{{B=0}}}, and we know that {{{C=k}}}, this means that {{{A=k}}} as well. But {{{A=4}}}. So {{{k=4}}}


This means that if {{{k=4}}}, then we get the circle  {{{4x^2 + 4y^2 - 8x + 17y -3=0}}} 



For any ellipse, just pick a positive 'k' value that is NOT equal to 4. This 'k' value will make {{{B^2-4AC<0}}} true.



For the parabola, just make {{{k=0}}} since this satisfies the equation {{{B^2-4AC=0}}} (basically, everything goes to zero since B and C are zero)



And finally, for any hyperbola, reverse the idea of the ellipse and pick any negative 'k' value. This works because {{{B^2-4AC>0}}} is essentially the opposite of {{{B^2-4AC<0}}}