Question

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

Answer 1

If it was y=x2, then (t,t2)? But how did you get t2−t+1? – ole dole May 10 '17 at 11:20

@oledole, distance of point (x0,y0) from line Ax+By+C=0 is

d=

|Ax0+By0+C|

√

A2+B2

Also, your curve is y2=x, hence a point on it is of the form (t2,t). – Galc127 May 10 '17 at 11:25

Then it makes sense, thanks – ole dole May 10 '17 at 11:34

Has another task there, almost the same just that: y=x2 and the line is x−y+1. Now the point on y=x2, will be (t,t2)? – ole dole May 10 '17 at 13:22

Answer 2

Both would have same slope at such condition.

For slope of:

x = y^2, 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

https://brainly.in/question/12672735