Question

Find the points on the curve y^2=4x which is nearest to the point (2,-8)

Answer 1

let (t^2,2t) is a point in given curve which is nearest from (2,-8)

S=distance between given points =root {(t^2-2)^2+(2t+8)^2}

S^2=(t^2-2)^2+(2t+8)^2
differentiate w.r.t t
2sds/dt=2 (t^2-2) 2t +2 (2t+8)2
ds/dt=0

t^3-2t+2t+8=0

t=-2

hence unknown point is (4,-4)

Answer 2

Answer:

The point (4, −4) on the curve $y^2=4x$ is nearest to the point (2, −8).

Step-by-step explanation:

To find : The points on the curve $y^2=4x$ which is nearest to the point (2,-8) ?

Solution :

The equation of the given curve is  $y^2=4x$

Let P (x, y) be a point on the curve, which is nearest to point A (2, −8).

Now, distance between the points A and P is given by:

$AP=\sqrt{(x-2)^2+(y+8)^2}$

$AP=\sqrt{(\frac{y^2}{4}-2)^2+(y+8)^2}$

$AP=\sqrt{(\frac{y^4}{16}-y^2+4)+(y^2+16y+64)}$

$AP=\sqrt{\frac{y^4}{16}+16y+68}$

Let  [tex]Z=AP^2=\frac{y^4}{16}+16y+68/tex]

Clearly, Z is maximum or minimum accordingly as AP is maximum or minimum. Also, P will be nearest to the point A if AP is minimum.

Now,  $\frac{dZ}{dy}=\frac{1}{16}\times 4y^2+16=\frac{y^3}{4}+16$

For maximum or minimum value of Z, we have

$\frac{dZ}{dy}=0$

$\frac{y^3}{4}+16=0$

$y^3+64=0$

 $(y+4)(y^2-4y+16)=0$

  $y=-4$

$\frac{d^2Z}{dy^2}=\frac{1}{4}\times 3y^2=\frac{3}{4}+y^2$

For y=-4,  $\frac{d^2Z}{dy^2}=\frac{3}{4}+(-4)^2=12>0$

Thus, Z is minimum when y = −4

Substituting y = −4 in the equation of the curve y2 = 4x, we obtain x = 4

Hence, the point (4, −4) on the curve $y^2=4x$ is nearest to the point (2, −8).