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The shortest distance between the line y-x=1 and curve x=y^2 is? Please give detailed answer and explanation Thanks

The shortest distance between the line y-x=1 and curve x=y^2 is?


Please give detailed answer and explanation


Thanks

Grade:upto college level

10 Answers

bhaveen kumar
38 Points
11 years ago

let us take a point on line y - x = 1 : A ( t -1 , t ) and on curve x = y2 : B ( t2 , t )

AB = root [ ( t - t )2 + (t2 - t + 1)2 ] = t2 - t + 1

f(x) = t2 - t + 1 , f '' (x) = 2t - 1 . For min value f '' (x) = 0 ; 2t - 1 =0 ; t =1/2

f '' '' (x) = 2 > 0 therefore t=1/2 is point of local minima

thus shortest distance = (1/2)2 - (1/2) + 1 = 3/4 units

Prajwal kr
49 Points
11 years ago

Any point on parabola, (k2,k)

Perpendicular distance formula:

D=(k-k2-1)/21/2

Differentiating and putting =0

1-2k=0

k=1/2

Therefore the point is (1/4, 1/2)

D=3/(321/2)

Yogita Bang
39 Points
11 years ago

The answer 3/√(32) is correct. THANKS for the correct answer.

But sir can you just tell me how that perpendicular distance formula is used. I would be really grateful to you.

Thanx once again

Prajwal kr
49 Points
11 years ago

Distance of any point(x1,y1) from a line ax+by+c is:

|ax1 + by1 + c|/(a2+b2)1/2

That is the formula. You can reply to this post if you want the derivation.

Jainam Shroff
26 Points
7 years ago
Please Provide The Derivation,Sir 
In Reply To

Distance of any point(x1,y1) from a line ax+by+c is:

 

|ax1 + by1 + c|/(a2+b2)1/2

 
That is the formula. You can reply to this post if you want the derivation.
Manonmani
11 Points
7 years ago
why cant we use the method used by bhaveen kumar...can i please know what is the mistake in it and why arent the answers matching???
Teju
11 Points
7 years ago
Mannon Mani, we cannot use that nethod because he took the points in both the curve and line with respect to t, which is incorrect as they are non intersecting. They cannot be taken with the same parameter
Yasaschandra Dvs
25 Points
7 years ago
what if i use the distance formulae and ill get s= mag( t-t^2-1/ sqrt 1+1)
=> s= t^2-t+1/sqrt(2) { since t^2-t+1 = (t-1/2) ^2+3 /4 >0}
and then ds/dt = 1/sqrt2 (2t-1)
&d^2 s / dt ^ 2= sqrt(2) >0
now ds/dt=0 => t=1/2
s is minimum at t=1/2 and the req shortest dist is(1/2)^2 – (1/2) +1 / sqrt(2)=3/4sqrt(2) => 3sqrt(2)/8
Samanvith
11 Points
6 years ago
Let (a2,a) be thw point of shortest distance on x=y2The distance between (a2,a) and line x−y+1=0 is given byD=a2−a+12–√=12–√[(a−12)2+34]If is min when a=12 and Dmin=342–√=>32–√8
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Any point on parabola, (k^2,k)
Perpendicular distance formula:
D = (k-k^2-1)/2^(1/2)
Differentiating and putting =0
1-2k = 0
k = 1/2
Therefore the point is (1/4, 1/2)
D = 3/(32^1/2)

Thanks and Regards

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