Question

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

Answer 1

Answer: 7/4

Step-by-step explanation:

Distance between two points is:

Sqrt((x1^2-x2^2)-(y1^2-y2^2)

If point A(x1,y1)E(y=x) => A(x;x)

If point B(x2;y2)E(y^2=x-2) => B(x;sqrt(x-2))

So the distance is:

Sqrt((x-x)^2+(x-sqrt(x-2)^2))=x-sqrt(x-2) *x-sqrt(x-2)>0*

Let f(x)=x-sqrt(x-2)

f'(x)=1-1/(2sqrt(x-2)

f'(x)=0 when x=9/4

So extremum(exactly minimum) of f(x) is x=9/4

f(9/4)=7/4

Shortest distance will be 7/4

Answer 2

Since solution to y = x and y^2 = x - 2 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 - 2, diff. wrt. x :

=> 2yy' = 1 => y' = 1/(2y)

For slope of line y = x, 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 - 2 => 9/4 = x

Hence the point where perpendicular meets on curve is (9/4, 1/2).

Now we need to find the point on line that is perpendicular to curve(or line with (9/4, 1/2) and slope = - 1/1).

Eqⁿ of perpendicular is

=> y - 1/2 = -1 (x - 9/4)

=> 4y + 4x - 11 = 0

Now, the lines y = x and 4y + 4x - 11 = 0 meet at :

(x, y) = (11/8, 11/8).

We have a point on line(11/8, 11/8) and perpendicularly a point(9/4, 1/2) on the curve. Using distance formula:

=> √(11/8 - 9/4)² + (11/8 - 1/2)²

=> (7√2)/8 or 7/(4√2) (both are same)