- Let
, and consider the two parabolas
![\[ y^2 = 4p(x-a), \qquad \text{and} \qquad x^2 = 4qy. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3a70e85bcd0d1c125b97704e8d8fe034_l3.png "Rendered by QuickLaTeX.com")
If these two parabolas are tangent to each other, show that the
-coordinate of the contact point depends only on 
.
- Find conditions for the constants
such that the two parabolas are tangent to each other.
Incomplete.
a) Parabolas being tangent to each other means that they have a common tangent line, so I would take first the derivative of the parabolas, to find the slope of any of their tangent lines. The slope of tangent lines to $P1$ is $y’ = \sqrt{\frac{p}{x-a}}$ and the tangent line slope of $P2$ is $y’ = \frac{x}{2q}$.
In order for them to have a common tangent line, their slopes must be equal so we can equate the last two equations to get: $y’ =\frac{x}{2q} = \sqrt{\frac{p}{x-a}} \implies \frac{p}{x-a} = \frac{x^2}{4q^2} \implies x^2(x-a) = 4pq^2$.
Actual tangent line is $y=y’x + m$, so at the point of contact $(x_0, y_0)$, we have $y_0 = \frac{x_0^2}{2q} + m = \sqrt{\frac{p}{x_0-a}} + m$ and $y_0^2=4p(x_0-a) \land 4y_0q=x_0^2$.
From this we get $m = -y_0$, and we can solve the equations to get $x_0 = \frac{4}{3}a$.
b) Plugging in the value of $x_0$ into the two tangent slopes, we get the condition $4a^3=27pq^2$