Analytical Geometry> shortest-distance...

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

Please give detailed answer and explanation

Thanks

Yogita Bang , 12 Years ago

Grade upto college level

10 Answers

bhaveen kumar

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

Last Activity: 12 Years ago

Prajwal kr

Any point on parabola, (k²,k)

Perpendicular distance formula:

D=(k-k²-1)/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)

Approved

Last Activity: 12 Years ago

Yogita Bang

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

Last Activity: 12 Years ago

Prajwal kr

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.

Last Activity: 12 Years ago

Jainam Shroff

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.

Last Activity: 9 Years ago

Manonmani

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???

Last Activity: 9 Years ago

Teju

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

Last Activity: 8 Years ago

Yasaschandra Dvs

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

Last Activity: 8 Years ago

Samanvith

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

Last Activity: 7 Years ago

Rishi Sharma

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

Last Activity: 5 Years ago