# Thread: Find equations of both lines

1. ## Find equations of both lines

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 $\displaystyle y={\sqrt{x} +3}$ 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.

2. 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:

$\displaystyle y = -x^2 - 1$

$\displaystyle \Rightarrow y' = -2x$

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 $\displaystyle y - y_1 = m(x - x_1)$

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

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

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

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

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

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

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

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

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

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

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

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

Tangent line 1:

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

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

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

Tangent line 2:

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

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

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

Below is a picture of what's going on

2. Find the point on the curve $\displaystyle y={\sqrt{x} +3}$ 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