The shortest distance between the line y=x and the curve y²=x-2 is:

Correct option is A. 742
We have, 2y.dy/dx = 1 ⇒ dy/dx = 1/2y
Let P(2+t², t) be a point on the curve y² = x - 2.
Then, the slope of the tangent at P is 1/2t.
The equation of the tangent at P is given by:
y - t = (1/2t)(x - (2+t²))
Since the shortest distance is along the normal, the slope of the normal is -2t.
The equation of the normal at P is:
y - t = -2t(x - (2+t²))
The line y = x intersects the normal at a point Q.
Substituting y = x in the equation of the normal:
x - t = -2t(x - (2+t²))
x - t = -2tx + 4t + 2t³
x(1+2t) = t(1+4+2t²)
x = t(2t²+5)/(2t+1)
y = x = t(2t²+5)/(2t+1)
The distance between P and Q is given by:
√((x - (2+t²))² + (y-t)²)
To minimize the distance, we differentiate with respect to t and set it to zero.
However, since the line y=x is perpendicular to the normal at the point of shortest distance:
The slope of the normal at P is -2t. The slope of the line y=x is 1.
Therefore, -2t = -1 => t = 1/2
Thus P = (9/4, 1/2)
So, shortest distance = |(9/4 - 1/2)/√2| = |7/4√2| = 7/(4√2) = 7√2/8 ≈ 742