Analytical Geometry> shortest-distance...

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

Yogita Bang , 11 Years ago

Grade upto college level

10 Answers

bhaveen kumar

Last Activity: 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

Last Activity: 11 Years ago

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

Yogita Bang

Last Activity: 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

Last Activity: 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

Last Activity: 8 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

Last Activity: 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

Last Activity: 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

Last Activity: 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

Last Activity: 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

Last Activity: 4 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