Question
Mathematics Question on applications of integrals
If P is a point on the parabola y=x2+4 which is closest to the straight line y=4x−1, then the co-ordinates of P are
A
(3,13)
B
(1,5)
C
(-2,8)
D
(2,8)
Answer
(2,8)
Explanation
Solution
P:y=x2+4 k=h2+4 L:y=4x−1 y−4x+1=0 d=AB=5k−4h+1 =5h2−4−4h+1 dhd(d)=52h−4=0 h=2 dh2d2(d)=52>0 ∴k=4+4=8 ∴ Point (2,8)