Question

Find the shortest distance between the line x - y + 1 = 0 and the curve $y^{2}$ = x

Answer 1

It's a bit tricky brother....

But I tried....

Is it correct???

Let (t2,t)(t2,t) be a point on y2=xy2=x. Its distance to the line x−y+1=0x−y+1=0is

∣∣∣t2−t+1(1)2+(−1)2−−−−−−−−−−√∣∣∣=(t−12)2+342–√|t2−t+1(1)2+(−1)2|=(t−12)2+342

So the shortest distance is 3/4√2.

Answer 2

Let (t2,t)(t2,t) be a point on y2=xy2=x.

Its distance to the line x−y+1=0x−y+1=0 is:-

=>t2−t+1(1)2+(−1)2

=>(t−12)2+342–√t2−t+1(1)2+(−1)2

=>(t−12)2+342

=>2t - 24 + 342

=> 2t = 318

=> t = 318/2 = 159
____________________
[Therefore, t = 318/2 = 159]
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[Answer:- 159]