Tangent Line to the curve that is Parallel to the line Find the equation of the tangent line to the curve $ y=1-x^2$ that is parallel to the line $2x-4y=8.$

I really don't know how to start this problem, can you help me please to solve this problem? Thank you.
Determine the slope of the given line: Since we want the tangent line to be parallel to the line $2x - 4y = 8$, we need for it to have the same slope. We can rewrite the equation in slope-intercept form to easily find slope: $4y= 2x- 8\iff y = \frac 12 x - 2$. Hence the desired slope of our line is $m = \frac 12$.
Then find the derivative of $y = 1-x^2$ and set it equal to the desired slope. $y' = -2x$. So put $-2x = \frac 12$ 
Solve for the value of $x$. $$-2x = \frac 12 \iff x = -\frac{1}{4}$$
That will give you the x-coordinate of a point on the desired line. Find the corresponding y-coordinate, and you'll have both coordinates of a point on the desired line. $$y = 1 - x^2 \implies y\left(-\frac 14\right) = 1 - \frac 1{16} = \frac {15}{16}$$
We know the slope of the desired line is equal to $\frac 12$, and we now have a point on the desired line, as well as on the given curve: i.e., the point of tangency $\left(-\frac 14, \frac {15}{16}\right)$.
Use the point-slope form to obtain the equation of the desired line. Given slope $m$ and a point on the desired line,$(x_0, y_0)$, the equation of the line is given by $$(y-y_0) = m(x-x_0)$$


This gives us: $$y - \frac {15}{16} = \frac 12\left(x + \frac 14\right) \iff 16y - 15= 8x + 2 \iff 8x - 16y= -17$$

We can graph the function, the given line, and the desired tangent line to see "what we've got". The "brownish-grey" line is the desired line:
The slope is not $0.5$ that is the slope for the line parallel to the line. The slope that should be used when solving its equation.
$y-y_1=m(x-x_1)$ the slope is 2 from $y= 2x$ which is the slope of the tangent line
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