Find the point P on the curve y2 = 4ax which is nearest to the point

(11a, 0).

the distance z from (x,y) to (11a,0) is

z = √((x-11a)^2 + y^2)
= √((x-11a)^2 + 4ax)
= √(x^2-18ax+121a^2)

dz/dx = (x-9a)/√(x^2-18ax+121a^2)
dz/dx=0 when x = 9a

So, P is (9a,6a)

To find the point P on the curve y^2 = 4ax that is nearest to the point (11a, 0), we can use the concept of the distance formula.

Step 1: Determine the general equation of the curve.
The given curve equation is y^2 = 4ax. This equation represents a parabola that opens to the right with the vertex at the origin (0,0).

Step 2: Calculate the distance between a general point P on the curve and the point (11a, 0).
Let's consider a general point P on the curve with coordinates (x, y). The distance formula between P and (11a, 0) is given by:

Distance = √[(x - 11a)^2 + (y - 0)^2]
= √[(x - 11a)^2 + y^2]

Step 3: Express y^2 in terms of x.
Since y^2 = 4ax, we can substitute this into the distance formula:

Distance = √[(x - 11a)^2 + 4ax]

Step 4: Find the minimum distance.
To find the minimum distance, we need to minimize the distance formula. This can be done by finding the value of x that minimizes the expression inside the square root.

To minimize √[(x - 11a)^2 + 4ax], we can take the derivative with respect to x and set it equal to zero:

d/dx [√((x - 11a)^2 + 4ax)] = 0

Using the chain rule and simplifying, we get:

(1/2) * (2(x - 11a) + 4a) / √((x - 11a)^2 + 4ax) = 0

Simplifying further, we have:

(x - 11a) + 2a = 0

x - 11a + 2a = 0

x = 9a

Step 5: Determine the y-coordinate of point P.
Substituting x = 9a back into the equation y^2 = 4ax, we can solve for y:

y^2 = 4a(9a) = 36a^2

y = ±6a

Since we are looking for the point nearest to (11a, 0), we take the positive y-coordinate:

y = 6a

Thus, the point P on the curve, nearest to (11a, 0), is (9a, 6a).

To find the point P on the curve y^2 = 4ax that is nearest to the point (11a, 0), we need to minimize the distance between these two points.

Let's denote the coordinates of point P as (x, y). Since the point lies on the curve y^2 = 4ax, we can substitute y^2 = 4ax into the distance formula to get the expression for the distance:

Distance = sqrt((x - 11a)^2 + y^2)

Now, we need to minimize this distance. To do that, we can take the derivative of the distance expression with respect to x, and set it equal to zero to find the critical points. Let's differentiate the distance expression:

d(Distance)/dx = d(sqrt((x - 11a)^2 + y^2))/dx

To find dy/dx, we can differentiate y^2 = 4ax with respect to x:

dy/dx = d(4ax)/dx
dy/dx = 4a

Now, substituting this value of dy/dx into the expression for the derivative of the distance expression:

d(Distance)/dx = d(sqrt((x - 11a)^2 + y^2))/dx
d(Distance)/dx = d(sqrt((x - 11a)^2 + (4a)^2))/dx
d(Distance)/dx = d(sqrt((x - 11a)^2 + 16a^2))/dx

To simplify the expression inside the square root, we can expand it:

(x - 11a)^2 + 16a^2 = x^2 - 22ax + 121a^2 + 16a^2
(x - 11a)^2 + 16a^2 = x^2 - 22ax + 137a^2

Now, taking the derivative of the simplified expression:

d(Distance)/dx = d(sqrt(x^2 - 22ax + 137a^2))/dx

Using the chain rule, the derivative of sqrt(u) is (1/2sqrt(u)) * du/dx:

d(Distance)/dx = (1/2sqrt(x^2 - 22ax + 137a^2)) * d(x^2 - 22ax + 137a^2)/dx
d(Distance)/dx = (1/2sqrt(x^2 - 22ax + 137a^2)) * (2x - 22a)

Setting this derivative equal to zero:

(1/2sqrt(x^2 - 22ax + 137a^2)) * (2x - 22a) = 0

Now we have two possibilities:

1. (1/2sqrt(x^2 - 22ax + 137a^2)) = 0

This implies that sqrt(x^2 - 22ax + 137a^2) = 0, but this cannot be true since the square root of a number cannot be zero.

2. (2x - 22a) = 0

Solving this equation for x:

2x - 22a = 0
2x = 22a
x = 11a

Now that we have the x-coordinate of the point P, we can substitute it back into the equation y^2 = 4ax to find the y-coordinate:

y^2 = 4ax
y^2 = 4a(11a)
y^2 = 44a^2
y = ±2√(11)a

Therefore, the two points on the curve y^2 = 4ax that are nearest to the point (11a, 0) are (11a, 2√(11)a) and (11a, -2√(11)a).