Find the shortest distance between,

$\displaystyle y = x^{2}-8x+15$

and

$\displaystyle 2y + 7 + 2x^{2} = 0$

We can say that the vector perpendicular to the tangent vector of each curve, $\displaystyle \vec{d}$, is described as follows,

$\displaystyle \vec{d} = a\hat{i} + b\hat{j} + c \hat{k}$

Now if I could get my two curves in the form of a position vector, I can differentiate both position vectors to obtain their tangent vectors.

Once I have their tangent vectors (Let's say vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$) I can cross them with my vector $\displaystyle \vec{d}$ and equate them to 0.

But how do I get my curves in the form of a position vector?

And even after all this, I've only got 2 equations, so I need another one! Where can I find that last equation?

EDIT:

$\displaystyle \vec{r_{1}(t)} = t\hat{i} + (t^{2} - 8t +15)\hat{j}$

$\displaystyle \vec{r_{2}(t)} = t\hat{i} + (-t^{2} - \frac{7}{2})\hat{j}$

How do those look for position vectors of my curves?