SOLUTION: 4x^2 + ky^2 - 8x + 17y = 3 Find the vaule of k to make this equation a circle, ellipse, hyperbola, and parabola; so different vaules for k that will make those different kinds o

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 4x^2 + ky^2 - 8x + 17y = 3 Find the vaule of k to make this equation a circle, ellipse, hyperbola, and parabola; so different vaules for k that will make those different kinds o      Log On


   



Question 308992: 4x^2 + ky^2 - 8x + 17y = 3
Find the vaule of k to make this equation a circle, ellipse, hyperbola, and parabola; so different vaules for k that will make those different kinds of equations.

Any single one of them would be helpful if you can't necessairly figure them all out, so if you know anything at all pleaseee help. Thank you so much!


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
This is very helpful to remember:


For the general conic Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0

If B%5E2-4AC%3C0, then the given conic above is an ellipse
Furthermore, if A=C, and B=0, then the conic is also a circle


If B%5E2-4AC=0, then the given conic above is a parabola


If B%5E2-4AC%3E0, then the given conic above is a hyperbola


First, let's subtract 3 from both sides to get 4x%5E2+%2B+ky%5E2+-+8x+%2B+17y+-3=0


So if we wanted to force 4x%5E2+%2B+ky%5E2+-+8x+%2B+17y+-3=0 to be a circle, then we must make sure that B%5E2-4AC%3C0, 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%5E2-4%284%29%28k%29%3C0 and simplify to get -16k%3C0. Solve for 'k' to get k%3E0. 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%5E2+%2B+4y%5E2+-+8x+%2B+17y+-3=0


For any ellipse, just pick a positive 'k' value that is NOT equal to 4. This 'k' value will make B%5E2-4AC%3C0 true.


For the parabola, just make k=0 since this satisfies the equation B%5E2-4AC=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%5E2-4AC%3E0 is essentially the opposite of B%5E2-4AC%3C0