# Thread: Find an equation of tangent

1. ## Find an equation of tangent

Find an equation of the line which is a tangent to both the parabola with equation y^2=4ax and the parabola with equation x^=4ay.

2. I've done this

3. Originally Posted by geton
I've done this
Why don't you share? Your solution might help someone else.

-Dan

4. Originally Posted by geton
Find an equation of the line which is a tangent to both the parabola with equation y^2=4ax and the parabola with equation x^=4ay.

Originally Posted by topsquark
Why don't you share? Your solution might help someone else.

-Dan
Let me try

$\displaystyle y^2 = 4ax \Rightarrow(x^2/4a)^2 = 4ax \Rightarrow x^4 = 64xa^3 \Rightarrow x = 0, \pm 4a$
Clearly -4a wont do. So the intersection points are (0,0),(4a,4a)

The tangent at (4a,4a)

$\displaystyle \frac{dy}{dx}\bigg{|}_ {x=4a} = \frac{d(\frac{x^2}{4a})}{dx}\bigg{|}_ {x=4a} = \frac{x}{2a}\bigg{|}_ {x=4a} = 2$

$\displaystyle y - 4a = \frac{dy}{dx}\bigg{|}_ {x=4a}(x - 4a) \Rightarrow y - 4a = 2(x - 4a) \Rightarrow y = 2x - 4a$

5. Originally Posted by topsquark
Why don't you share? Your solution might help someone else.

Let the tangent is y = mx + c

Substitute this into parabola's equation with y^2 = 4ax.
So: (mx + c)^2 = 4ax

There is only one root because tangent meets parabola at only one point.

From this equation we found mc = a

Same thing for x^2 = 4ay.

By solving two equations simultaneously, we get the tangent equation,
y + x +a = 0.