Given: point (4, 2) and parabola y^2 = 8x

To find: minimum distance from the point to the parabola

Solution:

Step 1: Write the equation of the parabola in terms of y

y^2 = 8x

x = y^2/8

Step 2: Find the equation of the tangent to the parabola at (x1, y1)

Differentiate x = y^2/8 with respect to y

dx/dy = y/4

Slope of tangent at (x1, y1) = dy/dx = 4/y1

Equation of tangent at (x1, y1) is given by:

(y - y1) = (4/y1)(x - x1)

Simplifying, we get:

y = (2y1/x1)(x + x1/2) - x1/2

Step 3: Find the point of intersection of the tangent and the line passing through (4, 2) and perpendicular to the tangent

Let the equation of the perpendicular line be y = mx + c

Slope of the perpendicular line = -1/slope of the tangent

= -y1/4

Using the point (4, 2), we get:

2 = -y1/4(4) + c

c = 2 + y1

Substituting the values of slope and y-intercept in the equation of the line, we get:

y = -y1/4(x - 4) + 2 + y1

Simplifying, we get:

y = (-y1/4)x + y1 + 3

Substituting this in the equation of the tangent, we get:

(-y1/4)x + y1 + 3 = (2y1/x1)(x + x1/2) - x1/2

Simplifying, we get:

(2y1/x1 - y1/4)x - y1 - x1/2 + 6 = 0

Comparing with the standard form of a straight line, we get:

a = 2y1/x1 - y1/4

b = -y1 - x1/2 + 6

c = 0

Step 4: Find the distance between (4, 2) and the point of intersection of the tangent and the line passing through (4, 2) and perpendicular to the tangent

Let the point of intersection be (x2, y2)

We have:

(x2 - 4)^2 + (y2 - 2)^2 = d^2

Substituting the values of a, b, and c, we get:

(x2 - 4)^2 + (y2 - 2)^2 = (b^2/a^2 + 1/a^2)(ax2 + by2 + c)^2

Simplifying, we get:

(x2 - 4)^2 + (y2 - 2)^2 = [(16x1^2 + 4y1^2 - 4x1y1)/x1^2][(2y1/x1 - y1/4)x2 - y1 - x1/2 + 6]^2

Substituting the value