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
To find the value of k such that the parabolas y^2 = 4x and y^2 = (x - k) have a common normal other than the x-axis, we can follow these steps:
Step 1: Understand the problem.
We have two parabolas given: y^2 = 4x and y^2 = (x - k). The common normal is a line that is perpendicular to both parabolas and touches them at the same point.
Step 2: Find the slope of the normals to the parabolas.
Since we are looking for a common normal, we need to find the slopes of the normals to both parabolas.
For the parabola y^2 = 4x, we can write it as y = ±√(4x), which means the slope of the normal will be the negative reciprocal of the derivative of √(4x) with respect to x. Differentiating, we get:
dy/dx = 2/√(4x).
The slope of the normal to this parabola is then -√(4x)/2, which simplifies to -√x.
For the parabola y^2 = (x - k), we can write it as y = ±√(x - k). Differentiating, we get:
dy/dx = 1 / (2√(x - k)).
The slope of the normal to this parabola is then -√(x - k) / (2√x - 2√k).
Step 3: Equate the slopes and solve for k.
Since the slopes of the normals to the parabolas should be equal (for a common normal), we can set the two expressions we found for the slopes equal to each other:
-√x = -√(x - k) / (2√x - 2√k).
To simplify this equation, we can square both sides to eliminate the square roots:
x = (x - k) / (4x - 4√(xk) + 4k).
Next, cross-multiply and simplify the equation:
4x^2 - 4kx = (x - k) * (4x - 4√(xk) + 4k).
4x^2 - 4kx = 4x^2 - 4√(xk)x + 4kx - 4k^2.
Canceling out like-terms:
-4kx = -4√(xk)x - 4k^2.
Since we are looking for the value of k, we can remove the x term by dividing both sides by -4x:
k = √(xk) + k^2 / x.
Simplifying further:
√(xk) + k^2 / x - k = 0.
Multiply through by x to eliminate the fraction:
x√(xk) + k^2 - kx = 0.
Rearranging the terms:
x√(xk) - kx + k^2 = 0.
Now, this equation is in terms of both x and k. To find a specific value of k, we need to substitute a value for x and solve for k.
Since we don't have any additional information or constraints, it's not possible to determine a specific value of k without more context. However, the equation x√(xk) - kx + k^2 = 0 defines the relationship between x and k for a common normal to the two parabolas.
2+1
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.
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
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.