Proving That $ \text{Spec}\Big( k[x,y,t]/(ty-x^{2}) \otimes_{k[t]} k (a) \Big) \cong \text{Spec}\Big( k[x,y]/(ay-x^{2}) \Big) $ $ k $ is an algebraically closed field, and $ a \in k. $ 
This question stems from Example 3.3.1 in Hartshorne's Algebraic Geometry. There is a surjective morphism $ f: \text{Spec}\Big(k[x,y,t]/(ty-x^{2}) \Big) \longrightarrow \text{Spec}(k[t]), $ and the closed points of $ k[t] $ are identified with elements of $ k $ because $ \text{Spec}(k[t]) = \mathbb{A}^{1}_{k}. $ I am trying to understand why the fibre $ X_a $ is the plane curve $ ay = x^{2} $ in $ \mathbb{A}^{2}_{k} $ and I think this is the same as establishing the isomorphism above.
 A: Intuituvely, the fiber $X_a$ corresponds with the curve associated to the point $t-a=0$, replacing we obtaing the required fact. But, algebraically, you are right since, writing $X=\mbox{Spec}\left(k[x,y,t]/(ty-x^2)\right)$, $Y=\mbox{Spec}\left(k[t]\right)$,
$$X_a=X\times_Y k(a)=\mbox{Spec}\left(k[x,y,t]/(ty-x^2)\right)\times_{k[t]}\mbox{Spec}\left(k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}\right)$$
and that is the spectrum of tha tensor product over $k[t]$, if we have an isomorphism between the global sections of affine schemes, we have that the schemes are isomophic, so that is enough to prove the isomorphism of that rings. 
Since $k$ is algebraically closed $k[t]_{(t-a)}/(t-a)\simeq k$ (think via evaluating in $a$ for intuition). Now, for the correspondence of rings that you require, consider the homomorphism 
$$\psi:k[x,y]\to k[x,y,t]/(yt-x^2)\otimes_{k[t]}k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$$
sending $p(x,y)\mapsto p(x,y)\otimes_{k[t]} 1$. It is clear that it is an homomorphism of rings. It is an epimorphism since if $\sum_i q_i(x,y,t)\otimes_{k[t]} a_i\in k[x,y,t]/(yt-x^2)\otimes_{k[t]}k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$, then
\begin{align}
\sum_i q_i(x,y,t)\otimes_{k[t]} a_i & = \sum_i a_i q_i(x,y,t)\otimes_{k[t]}1 \\
\text{with $a_i\in k$} \\
\sum_i q_i(x,y,t)\otimes_{k[t]} a_i & = \sum_j f_j(t)g_j(x,y)\otimes_{k[t]} 1 \\
& = \sum_j g_j(x,y)\otimes_{k[t]} f_j(t) \\
& = \sum_j g_j(x,y)\otimes_{k[t]} f_j(a) \\
& = \sum_j g_j(x,y)f_j(a)\otimes_{k[t]}1,
\end{align}
since $f_j(t)=f_j(a)$ in $k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$. Now, we have to prove that $\ker\psi=(ay-x^2)$. In fact, using the same trick (passing a polynomial over $t$ to the right side of the tensor product and evaluating on $a$) we get easily that $(ay-x^2)\subset\ker\psi$. Similar argument to the equality.
