Find the equation of the line tangent to y^2=-16x and parallel to x+y=1.
Please help me with this or just give me an idea on how to answer this problem... Thanks!
If the line is parallel to $\displaystyle x + y = 1$, then its gradient is $\displaystyle 1$. (Why?)
You want the equation of the tangent that has the same gradient.
So $\displaystyle y^2 = -16x$
$\displaystyle \frac{d}{dx}(y^2) = \frac{d}{dx}(-16x)$
$\displaystyle 2y\,\frac{dy}{dx} = -16$
$\displaystyle \frac{dy}{dx} = -\frac{8}{y}$.
So when the gradient is $\displaystyle 1$...
$\displaystyle -\frac{8}{y} = 1$
$\displaystyle -\frac{y}{8} = 1$
$\displaystyle y = -8$.
When $\displaystyle y = -8$ we have $\displaystyle (-8)^2 = -16x$
$\displaystyle 64 = -16x$
$\displaystyle x = -4$.
So we have the gradient $\displaystyle = 1$ and a coordinate $\displaystyle (-4, -8)$ that lies on the tangent.
Put this information into $\displaystyle y = mx + c$, solve for the unknown $\displaystyle c$, and then you will have the equation of the tangent.