The shortest distance between the line y-x=1 and curve x=y^2 is
Answers
Since solution to y - x = 1 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
Solution to a similar question 'shortest distance between a line y = x and curve y^2 = x - 2' https://brainly.in/question/12672735
