Find an equation for the line tangent to the curve at the point defined by the given value of $t$. Also, find the value of $\frac{d^{2}y}{d{x}^2}$ at this point.
$$x=8\cos t,y=4\sin t,t=\frac{\pi }{4}$$

Step 1 ：Given the parametric equations $x=8\cos t$ and $y=4\sin t$, we are asked to find the equation of the tangent line to the curve at the point defined by $t=\frac{\pi }{4}$ and the value of the second derivative of $y$ with respect to $x$ at this point.

Step 2 ：First, we find the derivative of $y$ with respect to $x$, which gives us the slope of the tangent line. We can find this derivative using the chain rule: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.

Step 3 ：Substituting the given equations into the chain rule, we get $\frac{dy}{dx}=\frac{4\cos (t)}{-8\sin (t)}=-\frac{\cos (t)}{2\sin (t)}$.

Step 4 ：Next, we find the second derivative of $y$ with respect to $x$ by taking the derivative of $\frac{dy}{dx}$ with respect to $t$, and then dividing by $\frac{dx}{dt}$. This gives us $\frac{d^{2}y}{d{x}^2}=-\frac{1/2+\cos (t{)}^2/(2\sin (t{)}^2)}{8\sin (t)}$.

Step 5 ：Substituting $t=\frac{\pi }{4}$ into the above equations, we get $\frac{dy}{dx}=-\frac{1}{2}$ and $\frac{d^{2}y}{d{x}^2}=-\frac{\sqrt{2}}{8}$.

Step 6 ：Substituting $t=\frac{\pi }{4}$ into the original parametric equations, we get the point $(x,y)=(4\sqrt{2},2\sqrt{2})$.

Step 7 ：Using the point-slope form of a line, $y-y_1=m(x-x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line, we can find the equation of the tangent line to be $y-2\sqrt{2}=-\frac{1}{2}(x-4\sqrt{2})$.

Step 8 ：Final Answer: The equation of the tangent line to the curve at the point defined by $t=\frac{\pi }{4}$ is $\boxed{{\displaystyle y-2\sqrt{2}=-\frac{1}{2}(x-4\sqrt{2})}}$. The value of the second derivative of $y$ with respect to $x$ at this point is $\boxed{{\displaystyle -\frac{\sqrt{2}}{8}}}$.