Step-by-step explanation:

Parabola: =>(x−1)2=−(y−1).....................(1) (a=41)

Line passing through (3,0) and (0,4) in intercept form =>3x+4y=1

                                                                                             =>4x+3y=12........................(2)

let there be a point on (1):A(1+2at,(1−at2))

=>A(1+t2,1−4t2)

Perpendicular distance between (A) and line (2),

=>L=∣∣∣∣∣∣∣∣∣16+94(1+2t)+3(1−4t2)−12∣∣∣∣∣∣∣∣∣

            =∣∣∣∣∣∣∣∣∣52t−43t2−5∣∣∣∣∣∣∣∣∣

            =∣∣∣∣∣∣58t+3t2−20∣∣∣∣∣∣

For minimum L, dtdL=0,−6t+8=0

=>t=34

So, L=∣∣∣∣∣∣15−44∣∣∣∣∣∣

       L=1544

Co ordinates of A:(1+64,1−4(3)16)

A:(35,3−1).

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