2x + x dy/dx + y = 0
dy/dx = (-2x - y)/x
from the tangent line we can see that the slope is 3
so,
(-2x - y)/x = 3
-2x - y = 3x
y = -5x
sub back into the original equation:
x^2 + x(-5x) + 16 = 0
x^2 - 5x^2 = -16
-4x^2 = -16
x^2 = 4
x = ± 2
when x = 2,
4 + 2y + 16 = 0
y = -10
so one of the tangents touches at (2,-10)
in y = 3x+k
-10 = 6+k
k = -16
you try for the other value of k, by letting x = -2
The line y=3x+k is a tagent to curve x^2+xy+16=0 find the possible value of k?
plz help me i don't no it
1 answer