Asked by Jatin
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
Answers
Answered by
Bosnian
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.
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.
Answered by
Steve
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
This should get you started...
https://math.stackexchange.com/questions/524130/common-normal-parabola-problem
Answered by
Steve
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.
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.
Answered by
Anonymous
2+1
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