Question

Find the point p on the curve y^2 =4ax,which is nearest to the point (11a,0)

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Answer 1

Given :

the curve y^2 =4ax

To find :

point nearest to the point (11a,0)

Let p (x,y)  be the nearest point

∴$D=\sqrt{(x-11a)^{2}+y^{2} }$

$S =(x-11a)^{2}$

$=(x-11a)^{2} +4ax$

$\frac{ds}{dx} =2(x-11a) +4a$                            ....as $( y^{2} =4ax)$

$\frac{ds}{dx} =0$ or $x= 9a$

∴ $y$ = ±$6a$

or $\frac{d^{2}s }{dx^{2} } =2$ ≥ $0$

for minimum distance, co ordinates are

P(9a, ± 6a)

Answer 2

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\: {y}^{2} = 4ax - - - (1)$

Let assume that the required point on the given curve is P(x, y) which is nearer to the point A(11a, 0).

Now, we know distance between two points A and P is given by

$\rm :\longmapsto\:AP = \sqrt{(x - 11a) ^{2} + {(y - 0)}^{2} }$

$\rm :\longmapsto\:AP = \sqrt{(x - 11a) ^{2} + {y}^{2} }$

$\rm :\longmapsto\:AP = \sqrt{(x - 11a) ^{2} + 4ax }$

On squaring both sides, we get

$\rm :\longmapsto\: {AP}^{2} = {(x - 11a)}^{2} + 4ax$

Let assume that

$\rm :\longmapsto\: f(x) = {(x - 11a)}^{2} + 4ax$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx} f(x) =\dfrac{d}{dx}[{(x - 11a)}^{2} + 4ax]$

$\rm :\longmapsto\:f'(x) = 2(x - 11a) + 4a$

For maxima and minima,

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:2(x - 11a) + 4a = 0$

$\rm :\longmapsto\:2x - 22a+ 4a = 0$

$\rm :\longmapsto\:2x - 18a = 0$

$\rm :\longmapsto\:2x = 18a$

$\bf\implies \:x = 9a$

Now, as we have

$\rm :\longmapsto\:f'(x) = 2(x - 11a) + 4a$

So, on differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:f''(x) = 2 > 0$

$\bf\implies \:AP \: is \: minimum \: when \: x = 9a$

On substituting the value of x = 9a, in equation (1), we get

$\rm :\longmapsto\: {y}^{2} = 4a(9a)$

$\rm :\longmapsto\: {y}^{2} = 36 {a}^{2}$

$\bf\implies \:y \: = \: \pm \: 6a$

So, required coordinates be (9a, 6a) or (9a, - 6a)

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Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.