Question

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

Answer 1

Answer:

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 2

Answer:

${\bf{\green{y^2 = 4 x}}}$

${\bf{\pink{4 x = y^2}}}$

${\bf{\blue{Divide both sides by 4}}}$

${\bf{\red{x = y^2/ 4}}}$

${\bf{\green{By using vertex form}}}$

${\bf{\orange{x = a ( y - k )^2+ h}}}$

${\bf{\blue{To get the values of a, h and k}}}$

${\bf{\pink{=>a=1/4}}}$

${\bf{\green{=>k=0}}}$

${\bf{\red{=>h=0}}}$

${\bf{\blue{Since the value of a is positive, the parabola opens right.}}}$

${\bf{\orange{Axis of symmetry: x = 0}}}$

${\bf{\green{Since parabolas have a common normal, axis of symmetry of prarabola}}}$

${\bf{\pink{ y^2= ( x - k ) also must be x = 0.}}}$

So,

${\bf{\blue{x - k = 0}}}$

${\bf{\red{0 - k = 0}}}$

${\bf{\green{Add k to both sides}}}$

${\bf{\orange{0 - k + k = 0 + k}}}$

${\bf{\blue{k = 0}}}$