find the point on the curve y^2 = 4x which is nearest to the point (2,-8) find the point on the curve y^2 = 4x which is nearest to the point (2,-8) @Mathematics

Question

zarkon · Answer

us the distance formula   (then take the derivative...set =0...yada yada yada)

anonymous · Answer

can't understand can you please explain with the steps!!!!

zarkon · Answer

plug your function and equation into the distance formula

$$D^2=(x_2-x_1)^2+(y_2-y_1)^2$$

anonymous · Answer

but how the equation is for a parabola

zarkon · Answer

you have two points (2,-8) and (x,y^2)

$$D^2=(x-2)^2+(y-(-8))^2$$

zarkon · Answer

oops   (4x,y^2)

anonymous · Answer

it should be (x, square root of 4x)

zarkon · Answer

write $$x=\frac{y^2}{4}$$

$$D^2=(x-2)^2+(y-(-8))^2=\left(\frac{y^2}{4}-2\right)^2+(y+8)^2$$

zarkon · Answer

differentiate with respect to y

zarkon · Answer

i get y=-4 and x=4

anonymous · Answer

|dw:1326017604265:dw|
see the drawing with due recognition to quadrants
we can see that focus lies (2,0)
so perependicular distance being the shortest , we will have some co-ordinate on the line drwn
which is of form (2,y)
put this into y^2=4ax
to get x and y
it is (2,2root2)