Parabola - find the equation

**lance** · February 22nd 2010, 10:27 PM

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!

**Prove It** · February 22nd 2010, 10:44 PM

Originally Posted by lance

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 $x + y = 1$ , then its gradient is $1$ . (Why?)

You want the equation of the tangent that has the same gradient.

So $y^2 = -16x$

$\frac{d}{dx}(y^2) = \frac{d}{dx}(-16x)$

$2y\,\frac{dy}{dx} = -16$

$\frac{dy}{dx} = -\frac{8}{y}$ .

So when the gradient is $1$ ...

$-\frac{8}{y} = 1$

$-\frac{y}{8} = 1$

$y = -8$ .

When $y = -8$ we have $(-8)^2 = -16x$

$64 = -16x$

$x = -4$ .

So we have the gradient $= 1$ and a coordinate $(-4, -8)$ that lies on the tangent.

Put this information into $y = mx + c$ , solve for the unknown $c$ , and then you will have the equation of the tangent.

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