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Question

Mathematics Question on applications of integrals

If PP is a point on the parabola y=x2+4y=x^{2}+4 which is closest to the straight line y=4x1y =4 x -1, then the co-ordinates of PP are

A

(3,13)

B

(1,5)

C

(-2,8)

D

(2,8)

Answer

(2,8)

Explanation

Solution

P:y=x2+4P: y=x^{2}+4 k=h2+4k=h^{2}+4 L:y=4x1L : y=4 x-1 y4x+1=0y-4 x+1=0 d=AB=k4h+15d=A B=\left|\frac{k-4 h+1}{\sqrt{5}}\right| =h244h+15=\left|\frac{h^{2}-4-4 h+1}{\sqrt{5}}\right| d(d)dh=2h45=0\frac{d(d)}{d h}=\frac{2 h-4}{\sqrt{5}}=0 h=2h=2 d2(d)dh2=25>0\frac{d^{2}(d)}{d h^{2}}=\frac{2}{\sqrt{5}}>0 k=4+4=8\therefore k=4+4=8 \therefore Point (2,8)