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Can you help me with a calculus question? I didn't manage to solve it.?
The parabola y = k - x^2 intersects the Y axis at point A, and the X axis at point B (point B is in the first quadrant). A tangent line touches the parabola at point C which is in the first quadrant. Point C's X-coordinate is 1. The line AB is parallel to the tangent line. Find k.
2 Answers
- ?Lv 74 years agoFavorite Answer
y = k − x²
y' = −2x
Point A has x-coordinate = 0 ----> A = (0, k)
Point B has y-coordinate = 0 ----> B = (√k, 0), k > 0
Point C has x-coordinate = 1 ----> tangent at C has slope = −2(1) = −2
AB is parallel to tangent at point C
Slope of AB = Slope of tangent at C
(k−0)/(0−√k) = √k2
−k/√k = −2
√k = 2
k = 4
- PinkgreenLv 74 years ago
y=k-x^2 is given
x=0=>y=k=>A=(0,k)
y=0=>x=sqr(k)=>B=(sqr(k),0)
y '(1)=-2(1)=-2
The slope of AB=-2
=>
(k-0)/(0-sqr(k))=-2
=>
k=2sqr(k)
=>
k^2-4k=0
=>
k(k-4)=0
=>
k=4 (k=/=0)
