Finding equation of a curve that a parametric equation of a line is tangent to

Question

I have the line $y=x\cdot\frac{-a}{-3a^2+4}$. I want to find the curve that this line is tangent to. More info: the tangency point needs to be at coordinates $(\frac{a\left(-3a^2+4\right)}{-9a^2+16},-\frac{a^2}{-9a^2+16})$

Clarifications:

The line should always be tangent to the curve, even when you plug in different values in the parameter.

Also, the curve must be constant, meaning that it should not be expressed via parameters.

Answer 1

the tangency point needs to be at coordinates $(\frac{a\left(-3a^2+4\right)}{-9a^2+16},-\frac{a^2}{-9a^2+16})$

The point of tangency is a point on your curve. So for each value of $a$, the point $$\left(\frac{a\left(-3a^2+4\right)}{-9a^2+16},-\frac{a^2}{-9a^2+16}\right)$$

is a point on your curve. I.e., a parametric equation for your curve is $$(x(t), y(t) ) =\left(\frac{t\left(-3t^2+4\right)}{-9t^2+16},-\frac{t^2}{-9t^2+16}\right)$$

If this curve doesn't meet your tangency condition (you can check that yourself), then you are out-of-luck. No curve can satisfy your conditions.