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Qn: the equation of the tangent to the curve y=kx+(6/x) at the point (-2,-19) is px + qy = c
find the values of k, p, q and c
2 answers
step 1
sub the point (-2,-19) into
y = kx - 6/x and solve for k (I got k=8)
step 2. Now that you know the actual function, find its derivative,
I got dy/dx = 8 - 6/x^2
sub in x=-2 to get the slope of the tangent, I got slope = 13/2
Step3. Now that you have the slope of the line and the point on the line,
find the equation.
I got 13x - 2y = c
sub in the point
-26 + 38 = c = 12
compare
13x - 2y = 12 with
px + qy = c
sub the point (-2,-19) into
y = kx - 6/x and solve for k (I got k=8)
step 2. Now that you know the actual function, find its derivative,
I got dy/dx = 8 - 6/x^2
sub in x=-2 to get the slope of the tangent, I got slope = 13/2
Step3. Now that you have the slope of the line and the point on the line,
find the equation.
I got 13x - 2y = c
sub in the point
-26 + 38 = c = 12
compare
13x - 2y = 12 with
px + qy = c
1 answer