Question

A straight line passes through (3, 0) and (0,4). A point A lies on the parabola $y=2x-x^2$ . Find the minimum distance from point A to the given straight line and also find the co-ordinates of A.

Answer 1

Parabola: $=>(x-1{)}^2=-(y-1).....................(1)$ $(a=\frac{1}{4})$

Line passing through $(3,0)$ and $(0,4)$ in intercept form $=>\frac{x}{3}+\frac{y}{4}=1$

$=>4x+3y=12........................\left(2\right)$

let there be a point on $\left(1\right):A(1+2at,(1-a{t}^2\left)\right)$

$=>A(1+\frac{2}{t},1-\frac{t^2}{4})$

Perpendicular distance between $\left(A\right)$ and line $\left(2\right)$,

$=>L=\left|\frac{4(1+\frac{t}{2})+3(1-\frac{t^2}{4})-12}{\sqrt{16+9}}\right|$

$=\left|\frac{2t-\frac{{3t}^2}{4}-5}{5}\right|$

$=\left|\frac{8t+3t^2-20}{5}\right|$

For minimum $L$, $\frac{dL}{dt}=0,-6t+8=0$

$=>t=\frac{4}{3}$

So, $L=\left|\frac{-44}{15}\right|$

$L=\frac{44}{15}$

Co ordinates of $A:(1+\frac{4}{6},1-\frac{16}{4\left(3\right)})$

$A:(\frac{5}{3},\frac{-1}{3}).$