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