# Thread: equation tanget of parabola

1. ## equation tanget of parabola

Find the equation of the common tangent to the parabolas y^2 = 4ax and x^2=4by

2. It is possible that there is an easier way to do this, but anyway:

Let the common tangent meet the first curve at $(x_1,y_1)$ and the second curve at $(x_2,y_2)$. Then the derivatives of those curves at those points are equal. Call this gradient m. then $y_1+m(x_2-x_1) = y_2$ (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.

3. ## Please tell me if what i have done is correct

Sir i have tried to solve this question as per ur directions .pls tell me if it is correct

Equation of 1st parabola is $y^2 = 4ax$
Let the point of tangent of the parabola be $(x_1,y_1)$
Therefore
Eq of tangent to the parabola will be $yy_1 = 2a(x+x_1)$

Similarly for the parabola $x^2=4by$
Eq of tangent to the 2nd parabola will be $xx_2 = 2b(y+y_2)$

Please tell me what needs to be done next ..........

4. Originally Posted by Shivanand
Find the equation of the common tangent to the parabolas y^2 = 4ax and x^2=4by

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 $y = 2\sqrt{a}\sqrt{x}$ and $y = \frac{1}{4b}x^2$.

Find their derivatives and set them equal to each other...

$\frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x$.

Solve for x.

$\frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x$

$2b\sqrt{a}= x^{\frac{3}{2}}$

$x = 2^{\frac{2}{3}}b^{\frac{2}{3}}a^{\frac{1}{3}}$.

Substitute this into one of the equations to find y.

Can you go from here?

5. ## Common tangents to parabolas

Hi -

Originally Posted by Prove It
Find their derivatives and set them equal to each other...

$\frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x$.

Solve for x.
Sorry, I'm afraid all this does is to find the value of $x$ - the same value on each curve - at which the tangents are parallel.

Can I suggest:

Re-write $x^2=4by$ as $y=\frac{x^2}{4b}$

$\therefore \frac{dy}{dx} = \frac{x}{2b}$

When $x=k$, $y=\frac{k^2}{4b}$ and $\frac{dy}{dx} = \frac{k}{2b}$

So the tangent to this curve at the point where $x = k$ is:

$y - \frac{k^2}{4b}= \frac{k}{2b}(x-k)$

i.e. $x=\frac{4by+k^2}{2k}$

This meets $y^2=4ax$ where

$y^2=2a \left(\frac{4by+k^2}{2k}\right)$

Re-arrange as a quadratic in $y$:

$ky^2-8aby-2ak^2=0$

Now this line is also a tangent to $y^2=4ax$ if this quadratic has equal roots.

So ...

(I'll leave you to do the next few steps.)

The equation turns out to be:

$b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3 }}=0$

6. ## Please tell if this is right

I have tried it the way u told me pls tell me if this is right

$y^2 = 4ax$ $\Rightarrow$[ math]y=2\sqrt{a}\sqrt{x}[/tex] $\rightarrow$(1)

$x^2=4by$ $\Rightarrow$ $y=\frac{x^2}{4b}$ $\rightarrow$(2)

Differetiating eq 1 wrt to x weget
$\frac{dy}{dx}$= $\frac{d}{dx}$ $(2\sqrt{a}\sqrt{x})$

= $2\sqrt{a}\frac{d}{dx}$ $(x)^\frac{1}{2}$

= $2\sqrt{a}.\frac{1}{2\sqrt{x}}$ $.\frac{d}{dx}$ $(x)$

= $\frac{\sqrt{a}}{\sqrt{x}}$ $(1)$

= $\frac{\sqrt{a}}{\sqrt{x}}$ $\rightarrow(3)$

Similarly for eq 2

$\frac{dy}{dx}=$ $\frac{d}{dx}$ $[\frac{x^2}{4b}]$

= $\frac{1}{4b}$ $\frac{d}{dx}$ $x^2$

= $\frac{1}{4b}$ $2x$

= $\frac{x}{2b}$ $\rightarrow(4)$

Setting them equal to each other weget

$\frac{\sqrt{a}}{\sqrt{x}}$= $\frac{x}{2b}$

$2b\sqrt{a}$= $x\sqrt{x}$

$2b\sqrt{a}$= $x.x^\frac{1}{2}$

$2b\sqrt{a}$= $x^\frac{3}{2}$

$2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{2}$ $^\frac{2}{3}$= $x$

$2^\frac{2}{3}$ $b^\frac{2}{3}$ $a^\frac{1}{3}$=x

Therefore x = $2^\frac{2}{3}$ $b^\frac{2}{3}$ $a^\frac{1}{3}$

Now substituting x in one of the equation $y^2=4ax$ weget

$y^2=4ax$

$y^2$ = $4a(2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})$

y= $\sqrt{4}\sqrt{2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{ 4}{3}}$

y= $2(2^\frac{2}{3} .b^\frac{2}{3}.a^\frac{4}{3})^\frac{1}{2}$

y= $2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})$

Now the equation of tangent to parabola $y^2=4ax$ at point $(x_1,y_1)$ would be
$yy_1=2a(x+x_1)$

ie
$y.2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})$= $2a(x+2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})$

7. ## Is it right?

Hi-

Originally Posted by varunnayudu
pls tell me if this is right
No, sorry, it's not right at all. Please see my previous posting.

8. ## Hw did u get this answer.

i.e.

This meets where

sir hw did u get this answer

when $y^2 = 4ax$

and $x = \frac{4by+k^2}{2k}$

$
y^2 = 4a (\frac{4by+k^2}{2k})
$

Hw did u get

9. ## Typo!

Hello varunnayudu
Originally Posted by varunnayudu
sir hw did u get this answer

when $y^2 = 4ax$

and $x = \frac{4by+k^2}{2k}$

$
y^2 = 4a (\frac{4by+k^2}{2k})
$

Hw did u get

Oops! Typo. I copied and pasted, and forgot to remove the 2 from the denominator. Of course, it should read:

$
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

$(-8ab)^2 = 4k(-2ak^2)$

$\implies -8ak^3 = 8^2a^2b^2$

$\implies k^3=-8ab^2$

$\implies k=-2a^{\frac{1}{3}}b^{\frac{2}{3}}$

Re-arrange the equation of the tangent found previously to read:

$4by -2kx + k^2=0$

and then substitute this value of $k$ into this equationto get:

$4by +4a^{\frac{1}{3}}b^{\frac{2}{3}}x + 4a^{\frac{2}{3}}b^{\frac{4}{3}}=0$

i.e. $b^{\frac{1}{3}}y+a^{\frac{1}{3}}x + (ab)^{\frac{2}{3}}=0$

I hope that you understand this now.

10. ## Hw did u get this

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 $y =-8ab$ and $x = -2ak^2$

specify this

11. ## And Hw did u get this

Re-arrange the equation of the tangent found previously to read:

----------------------------------------

which eq do i have to substitute to get that eq as stated above

12. ## Equal roots of a quadratic

Hello varunnayudu
Originally Posted by varunnayudu
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 $y =-8ab$ and $x = -2ak^2$

specify this

$ax^2 +bx+c=0$ has equal roots if $b^2=4ac$.

So

has equal roots if:

$(-8ab)^2 = 4k(-2ak^2)$

13. ## Equation of common tangent

Hello varunnayudu
Originally Posted by varunnayudu

Originally Posted by varunnayudu
Re-arrange the equation of the tangent found previously to read:

----------------------------------------

which eq do i have to substitute to get that eq as stated above

Here's a summary of how we solved this question.

By differentiation, we found that the gradient of $x^2=4by$
when $x=k$ is $\frac{k}{2b}$, and the value of $y$ at this point is $\frac{k^2}{4b}$.

Then, using $y - y_1 = m(x-x_1)$
we found the tangent to $x^2=4by$ at the point $(k, \frac{k^2}{4b})$ is

$y-\frac{k^2}{4b}=\frac{k}{2b}(x-k)$

Re-arranging and simplifying this equation we get:

$4by -2kx +k^2=0$ (1)

Next, we found the value of $k$
that made this line a tangent to $y^2=4ax$, using the fact that a tangent meets a curve at two coincident points, and therefore the quadratic equation giving the values of $y$ has equal roots.

Then into equation (1) we substitute this value of $k$ (= $-2a^{\frac{1}{3}}b^{\frac{2}{3}}$)
to obtain the tangent that is common to both parabolas. When we do this, and simplify the result, we get

$b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3 }}=0$