
Steve S. answered 12/15/13
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
Find the y-intercept of the tangent line to the curve y=sqrt(x^2+33). The answer is 33/7.
You neglected to tell us what point on the curve the tangent line intersects. Since you did give us the y-intercept of the tangent line, let's find the point(s).
y=sqrt(x^2+33)
y' = x/sqrt(x^2+33) = x/y
Assume the point(s) on the curve where the tangent line(s) intersect are (p,q).
q= y(p) = sqrt(p^2+33)
Equation of tangent line(s) is: y - q = (p/q) * (x - p)
Or: y = (p/q) * x + q - (p/q) * p
The y-intercept = q - p^2/q = 33/7
(q^2 - p^2)/q = 33/7
(p^2 + 33 - p^2)/sqrt(p^2+33) = 33/7
Numerators are the same so denominators are equal:
sqrt(p^2+33) = 7
Square both sides:
p^2 + 33 = 49
p^2 = 16
p = ±4
q = sqrt((±4)^2+33) = 7
So tangent lines at either of the two points (±4,7) have the same y-intercept = 33/7.


Steve S.
At the top the student wrote, "The answer is 33/7."
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