If the two parabolas y^2=4x and y2=(x -k) have a common normal other than the x-axis then k can be equal to

Answer 1

y² = 4 x

4 x = y²

Divide both sides by 4

x = y² / 4

Use the vertex form:

x = a ( y - k )² + h

to determine the values of a, h and k.

a = 1 / 4 , k = 0 , h = 0

Since the value of a is positive, the parabola opens right.

Axis of symmetry: x = 0

Since parabolas have a common normal, axis of symmetry of prarabola y² = ( x - k ) also must be x = 0.

So:

x - k = 0

0 - k = 0

Add k to both sides

0 - k + k = 0 + k

0 = k

k = 0

Equation of parabola:

y² = ( x - k )

y² = ( x - 0 )

y² = x

If you want in wolframalpha.c o m type:

plot y^2=4x , y^2=x

and clix option =

you will see graph.

Answer 2

Eh? How can two parabolas with a common vertex and a common axis of symmetry have two distinct normals?

This should get you started...

https://math.stackexchange.com/questions/524130/common-normal-parabola-problem

Answer 3

I'll work with a vertical axis, just because it makes me more comfortable with the derivatives.

For y = x^2/4 (x^2 = 4y), at x=a, the slope is x/2, so the normal has slope -2/a

For y = x^2+k, at x=b, the slope is 2b, so the normal has slope -1/(2b)

So, the equations of our normal lines are

y = -2/a (x-a) + a^2/4
y = -1/(2b) (x-b) + b^2+k
or,

y = -2/a x + 2 + a^2/4
y = -1/(2b) x + b^2 + k + 1/2

We want those two lines to be the same, so working in the 1st quadrant (so a and b are both positive), that means that

1/b = 2/a
a = 4b

1/2 + b^2 + k = 2 + a^2/4
3b^2 = k - 3/2

So, pick any value for k > -3/2 and you can find the equation of the normal line common to both parabolas.

Answer 4

2+1