Find equations of both lines

**johntuan** · June 24th 2007, 08:38 AM

1.Find the equation of both lines that pass through the point (-2,4) and are tangent to the the parabola y=-x^2 -1. Show all work.

2. Find the point on the curve $y={\sqrt{x} +3}$ where the tangent line is perpendicular to the line 6x+y-5=0.

Can someone help me with this please and list the steps in solving this problem.

**Jhevon** · June 24th 2007, 09:20 AM

Originally Posted by johntuan

1.Find the equation of both lines that pass through the point (-2,4) and are tangent to the the parabola y=-x^2 -1. Show all work.

First thing's first, let's find the formula for the slope of any tangent line to the curve, we get that from the derivative:

$y = -x^2 - 1$

$\Rightarrow y' = -2x$

So every tangent to the curve will have the slope -2x. So use that for our m.

Now we have by the point-slope form, that $y - y_1 = m(x - x_1)$

Using $(x_1, y_1) = (-2,4)$ and $m = -2x$ , we get:

$y - 4 = -2x(x + 2)$

$\Rightarrow y = -2x^2 - 4x + 4$ ------> equation for any tangent to the curve

Now, want all lines that intersect with the curve $y = -x^2 - 1$ , so we equate the general formula for the tangent line to the formula for the curve, we get:

$-2x^2 - 4x + 4 = -x^2 - 1$

$\Rightarrow x^2 + 4x - 5 = 0$

$\Rightarrow (x + 5)(x - 1) = 0$

$\Rightarrow x = -5 \mbox { or } x = 1$

Using each of these values, we can get a different m

when $x = -5$ , $m_1 = 10$

when $x = 1$ , $m_2 = -2$

Again going back to the point-slope form, using $(x_1,y_1) = (-2,4)$ and $m_1 = 10$ and $m_2 = -2$ , we get:

Tangent line 1:

$y - y_1 = m_1 (x - x_1)$

$\Rightarrow y - 4 = 10(x + 2)$

$\Rightarrow \boxed {y = 10x + 24}$

Tangent line 2:

$y - y_1 = m_2 (x - x_1)$

$\Rightarrow y - 4 = -2(x + 2)$

$\Rightarrow \boxed {y = -2x}$

Below is a picture of what's going on

2. Find the point on the curve $y={\sqrt{x} +3}$ where the tangent line is perpendicular to the line 6x+y-5=0.

Questions very similar to this were done yesterday. See if you can read through the threads and understand what was done. it's not that hard to understand, you're bound to get one of the explanations

http://www.mathhelpforum.com/math-he...d-tangent.html

http://www.mathhelpforum.com/math-he...gent-line.html

Thread: Find equations of both lines

LinkBack

Thread Tools

Display

Find equations of both lines

Similar Math Help Forum Discussions

find equations of tan lines parallel to line

Find equations of the tangent lines at points

Find the values of k so that lines are perpendicular using symetric equations

Equations of Lines

lines an equations

Search Tags