Question

Part A: Find the point on the parabola y^2 = 2x that is closest to point (1, 4)
which I got to be (2, root(5) )

then I need help with
PART B: Verify that the tangent to the parabola y^2 = 2x at your answer is perpendicular to the line connecting your answer to the point (1, 4)

could someone just explain to me the process of Part B?

Answer 1

let (x,y)be the point close to (1,4)
then distance
\[d=\sqrt{\left( x-1 \right)^{2}+\left( y-4 \right)^{2}}\]
now (x1,y1) lies on parabola
\[(y)^{2}=2x,x=\frac{ y ^{2} }{ 2 }\]
\[d=\sqrt{\left( \frac{ y ^{2} }{ 2 }-1 \right)^{2}+\left( y-4 \right)^{2}}\]
Let \[D=d ^{2}=\left( \frac{ y ^{2} }{ 2 }-1 \right)^{2}+\left( y-4 \right)^{2}\]
\[\frac{ dD }{ dy }=2\left( \frac{ y ^{2} }{ 2 } -1\right)\frac{ 2y }{ 2 }+2(y-4)\]
\[=y ^{3}-2y+2y-8\]
\[=y ^{3}-8\]
\[\frac{ dD }{ dy }=0 gives y ^{3}-8=0,y=2\]
\[\frac{ d ^{2}D }{ dy ^{2} }=3y ^{2}\]
\[at~y=2,\frac{ d ^{2}D }{ dy ^{2} }=3*2^{2}=12>0\]
hence D is minimum at y=2
\[2x=y ^{2}=2^{2}=4,x=2\]
Hence point nearest to (1,4) is (2,2)

Answer 2

find dy/dx at (2,2) from y^2=2x
this is the slope of tangent
also find the slope of line through (1,4) and (2,2)
multiply them . if product is-1 then they are perpendicular.

Answer 3

correction above write only (x,y) in place of (x1,y1)