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 $\displaystyle (x_1,y_1)$ and the second curve at $\displaystyle (x_2,y_2)$. Then the derivatives of those curves at those points are equal. Call this gradient m. then $\displaystyle y_1+m(x_2-x_1) = y_2$ (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 $\displaystyle y^2 = 4ax $
Let the point of tangent of the parabola be $\displaystyle (x_1,y_1)$
Therefore
Eq of tangent to the parabola will be $\displaystyle yy_1 = 2a(x+x_1)$
Similarly for the parabola $\displaystyle x^2=4by$
Eq of tangent to the 2nd parabola will be $\displaystyle xx_2 = 2b(y+y_2)$
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 $\displaystyle y = 2\sqrt{a}\sqrt{x}$ and $\displaystyle y = \frac{1}{4b}x^2$.
Find their derivatives and set them equal to each other...
$\displaystyle \frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x$.
Solve for x.
$\displaystyle \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x$
$\displaystyle 2b\sqrt{a}= x^{\frac{3}{2}}$
$\displaystyle x = 2^{\frac{2}{3}}b^{\frac{2}{3}}a^{\frac{1}{3}}$.
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 $\displaystyle x$ - the same value on each curve - at which the tangents are parallel.
Can I suggest:
Re-write $\displaystyle x^2=4by$ as $\displaystyle y=\frac{x^2}{4b}$
$\displaystyle \therefore \frac{dy}{dx} = \frac{x}{2b}$
When $\displaystyle x=k$, $\displaystyle y=\frac{k^2}{4b}$ and $\displaystyle \frac{dy}{dx} = \frac{k}{2b}$
So the tangent to this curve at the point where $\displaystyle x = k$ is:
$\displaystyle y - \frac{k^2}{4b}= \frac{k}{2b}(x-k)$
i.e. $\displaystyle x=\frac{4by+k^2}{2k}$
This meets $\displaystyle y^2=4ax$ where
$\displaystyle y^2=2a \left(\frac{4by+k^2}{2k}\right)$
Re-arrange as a quadratic in $\displaystyle y$:
$\displaystyle ky^2-8aby-2ak^2=0$
Now this line is also a tangent to $\displaystyle y^2=4ax$ if this quadratic has equal roots.
So ...
(I'll leave you to do the next few steps.)
The equation turns out to be:
$\displaystyle b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3 }}=0$
Grandad
I have tried it the way u told me pls tell me if this is right
$\displaystyle y^2 = 4ax$ $\displaystyle \Rightarrow$[ math]y=2\sqrt{a}\sqrt{x}[/tex]$\displaystyle \rightarrow$(1)
$\displaystyle x^2=4by$ $\displaystyle \Rightarrow$ $\displaystyle y=\frac{x^2}{4b}$$\displaystyle \rightarrow$(2)
Differetiating eq 1 wrt to x weget
$\displaystyle \frac{dy}{dx}$= $\displaystyle \frac{d}{dx}$$\displaystyle (2\sqrt{a}\sqrt{x})$
=$\displaystyle 2\sqrt{a}\frac{d}{dx}$$\displaystyle (x)^\frac{1}{2}$
=$\displaystyle 2\sqrt{a}.\frac{1}{2\sqrt{x}}$$\displaystyle .\frac{d}{dx}$$\displaystyle (x)$
=$\displaystyle \frac{\sqrt{a}}{\sqrt{x}}$$\displaystyle (1)$
=$\displaystyle \frac{\sqrt{a}}{\sqrt{x}}$ $\displaystyle \rightarrow(3)$
Similarly for eq 2
$\displaystyle \frac{dy}{dx}=$$\displaystyle \frac{d}{dx}$$\displaystyle [\frac{x^2}{4b}]$
=$\displaystyle \frac{1}{4b}$$\displaystyle \frac{d}{dx}$$\displaystyle x^2$
=$\displaystyle \frac{1}{4b}$$\displaystyle 2x$
=$\displaystyle \frac{x}{2b}$$\displaystyle \rightarrow(4)$
Setting them equal to each other weget
$\displaystyle \frac{\sqrt{a}}{\sqrt{x}}$=$\displaystyle \frac{x}{2b}$
$\displaystyle 2b\sqrt{a}$=$\displaystyle x\sqrt{x}$
$\displaystyle 2b\sqrt{a}$=$\displaystyle x.x^\frac{1}{2}$
$\displaystyle 2b\sqrt{a}$=$\displaystyle x^\frac{3}{2}$
$\displaystyle 2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{2}$$\displaystyle ^\frac{2}{3}$=$\displaystyle x$
$\displaystyle 2^\frac{2}{3}$$\displaystyle b^\frac{2}{3}$$\displaystyle a^\frac{1}{3}$=x
Therefore x = $\displaystyle 2^\frac{2}{3}$$\displaystyle b^\frac{2}{3}$$\displaystyle a^\frac{1}{3}$
Now substituting x in one of the equation $\displaystyle y^2=4ax$ weget
$\displaystyle y^2=4ax$
$\displaystyle y^2$ =$\displaystyle 4a(2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})$
y=$\displaystyle \sqrt{4}\sqrt{2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{ 4}{3}}$
y=$\displaystyle 2(2^\frac{2}{3} .b^\frac{2}{3}.a^\frac{4}{3})^\frac{1}{2}$
y=$\displaystyle 2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})$
Now the equation of tangent to parabola $\displaystyle y^2=4ax$ at point $\displaystyle (x_1,y_1)$ would be
$\displaystyle yy_1=2a(x+x_1)$
ie
$\displaystyle y.2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})$=$\displaystyle 2a(x+2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})$
Hello varunnayudu
Oops! Typo. I copied and pasted, and forgot to remove the 2 from the denominator. Of course, it should read:
$\displaystyle
y^2 = 2a (\frac{4by+k^2}{k})
$
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
$\displaystyle (-8ab)^2 = 4k(-2ak^2)$
$\displaystyle \implies -8ak^3 = 8^2a^2b^2$
$\displaystyle \implies k^3=-8ab^2$
$\displaystyle \implies k=-2a^{\frac{1}{3}}b^{\frac{2}{3}}$
Re-arrange the equation of the tangent found previously to read:
$\displaystyle 4by -2kx + k^2=0$
and then substitute this value of $\displaystyle k$ into this equationto get:
$\displaystyle 4by +4a^{\frac{1}{3}}b^{\frac{2}{3}}x + 4a^{\frac{2}{3}}b^{\frac{4}{3}}=0$
i.e. $\displaystyle b^{\frac{1}{3}}y+a^{\frac{1}{3}}x + (ab)^{\frac{2}{3}}=0$
I hope that you understand this now.
Grandad
Sir,
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
---------------
Hw did u substitute $\displaystyle y =-8ab$ and $\displaystyle x = -2ak^2 $
specify this
Hello varunnayudu
Here's a summary of how we solved this question.
By differentiation, we found that the gradient of $\displaystyle x^2=4by$ when $\displaystyle x=k$ is $\displaystyle \frac{k}{2b}$, and the value of $\displaystyle y$ at this point is $\displaystyle \frac{k^2}{4b}$.
Then, using $\displaystyle y - y_1 = m(x-x_1)$ we found the tangent to $\displaystyle x^2=4by$ at the point $\displaystyle (k, \frac{k^2}{4b})$ is
$\displaystyle y-\frac{k^2}{4b}=\frac{k}{2b}(x-k)$
Re-arranging and simplifying this equation we get:
$\displaystyle 4by -2kx +k^2=0$ (1)
Next, we found the value of $\displaystyle k$ that made this line a tangent to $\displaystyle y^2=4ax$, using the fact that a tangent meets a curve at two coincident points, and therefore the quadratic equation giving the values of $\displaystyle y$ has equal roots.
Then into equation (1) we substitute this value of $\displaystyle k$ (= $\displaystyle -2a^{\frac{1}{3}}b^{\frac{2}{3}}$) to obtain the tangent that is common to both parabolas. When we do this, and simplify the result, we get
$\displaystyle b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3 }}=0$
No more questions about this problem now, please.
Grandad