Find the equation of the common tangent to the parabolas y^2 = 4ax and x^2=4by
It is possible that there is an easier way to do this, but anyway:
Let the common tangent meet the first curve at and the second curve at . Then the derivatives of those curves at those points are equal. Call this gradient m. then (draw a diagram if you are unsure of this one). This gives
5 equations for 5 unknowns, which surprisingly solve quite easily. Remember that you only need one point and m to find the equation of the tangent.
Sir i have tried to solve this question as per ur directions .pls tell me if it is correct
Equation of 1st parabola is
Let the point of tangent of the parabola be
Therefore
Eq of tangent to the parabola will be
Similarly for the parabola
Eq of tangent to the 2nd parabola will be
Please tell me what needs to be done next ..........
The common tangent is found by the derivatives being equal.
So we'd need to write both equations as y in terms of x.
The equations are and .
Find their derivatives and set them equal to each other...
.
Solve for x.
.
Substitute this into one of the equations to find y.
Can you go from here?
Hi -
Sorry, I'm afraid all this does is to find the value of - the same value on each curve - at which the tangents are parallel.
Can I suggest:
Re-write as
When , and
So the tangent to this curve at the point where is:
i.e.
This meets where
Re-arrange as a quadratic in :
Now this line is also a tangent to if this quadratic has equal roots.
So ...
(I'll leave you to do the next few steps.)
The equation turns out to be:
Grandad
I have tried it the way u told me pls tell me if this is right
[ math]y=2\sqrt{a}\sqrt{x}[/tex] (1)
(2)
Differetiating eq 1 wrt to x weget
=
=
=
=
=
Similarly for eq 2
=
=
=
Setting them equal to each other weget
=
=
=
=
=
=x
Therefore x =
Now substituting x in one of the equation weget
=
y=
y=
y=
Now the equation of tangent to parabola at point would be
ie
=
Hello varunnayudu
Oops! Typo. I copied and pasted, and forgot to remove the 2 from the denominator. Of course, it should read:
The next line of my posting is correct, though. Re-arrange as a quadratic in :
Let me complete the working for you.
Now this line is also a tangent to if this quadratic has equal roots; i.e. if
Re-arrange the equation of the tangent found previously to read:
and then substitute this value of into this equationto get:
i.e.
I hope that you understand this now.
Grandad
Hello varunnayudu
Here's a summary of how we solved this question.
By differentiation, we found that the gradient of when is , and the value of at this point is .
Then, using we found the tangent to at the point is
Re-arranging and simplifying this equation we get:
(1)
Next, we found the value of that made this line a tangent to , using the fact that a tangent meets a curve at two coincident points, and therefore the quadratic equation giving the values of has equal roots.
Then into equation (1) we substitute this value of (= ) to obtain the tangent that is common to both parabolas. When we do this, and simplify the result, we get
No more questions about this problem now, please.
Grandad