Find the possible values of k in the case where x = -6 is a tangent to C.
I spent so long on this question and I give up because I keep going around in circles and I don't know what to do
If it was any equation that didn't lead to a vertical line with infinite gradient, then I could've solved it so much quicker.
I think I know!
After literally 30 minutes of thinking, the idea occured to me:
Rather than letting dy/dx = infinity, I can simply let dx/dy = 0.
Actually I don't know if that helps me with this specific question
It would've helped me in the last one though.
OMFG yes I finally found the answer after almost an hour!
Let dx/dy = 0
(6y^2 + x) / (-3x^2 - y) = 0
6y^2 + x = 0
y^2 = -x/6
y = +or- sqrt(-x/6)
when x = -6
y = 1 or y = -1
If y = 1:
216 + 4 + k = 0 so k = -220
If y = -1:
216 - 4 + k = 0 so k = -212
So k = -220 or k = -212