Question

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

Answer 1

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 + 1f(x) = t2 - t + 1 , f '' (x) = 2t - 1 . For min value f '' (x) = 0 ;2t - 1 =0 ; t =1/2f '' '' (x) = 2 > 0 therefore t=1/2 is point of local minimathus shortest distance = (1/2)2 - (1/2) + 1 = 3/4 units.

Answer 2

Step-by-step explanation:

Since solution to x - y + 1 = 0 and y^2 = x doesn't exist, the shortest distance between them must be perpendicular to the curve and line. It means, both would have same slope at such condition.

For slope of:

y^2 = x, differen. wrt. x => y' = 1/(2y)

y - x = 1, use y = mx + c => m = 1

As both should have same slope, m = y'

=> 1 = 1/2y => y = 1/2

Therefore, when y = 1/2 (on the curve)

(1/2)^2 = x => 1/4 = x

Hence the point where perpendicular meets on curve is (1/4, 1/2).

Now we need to find the point on line that is perpendicular to curve(or line with (1/4, 1/2) and slope = - 1/1).

Eqⁿ of perpendicular is

=> y - 1/2 = -1 (x - 1/4)

=> 4y + 4x - 3 = 0

Now, the lines y - x = 1 and 4y + 4x - 3 = 0 meet at: (x, y) = (-1/8, 7/8).

We have a point on line(-1/8, 7/8) and perpendicularly a point(1/4, 1/2) on the curve. Using distance formula:

=> √(11/8 - 9/4)² + (11/8 - 1/2)²

=> 3√2/8

Read an answer to a similar question:

shortest distance between a line y = x and curve y^2 = x - 2

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