Results 1 to 2 of 2

Math Help - Find equations of both lines

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    120

    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 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.
    Last edited by johntuan; June 24th 2007 at 09:00 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by johntuan View Post
    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
    Attached Thumbnails Attached Thumbnails Find equations of both lines-tangents.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find equations of tan lines parallel to line
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 18th 2014, 02:31 PM
  2. Find equations of the tangent lines at points
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 27th 2009, 10:50 PM
  3. Replies: 1
    Last Post: April 27th 2009, 05:20 AM
  4. Equations of Lines
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 17th 2008, 06:20 PM
  5. lines an equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 14th 2006, 09:21 AM

Search Tags


/mathhelpforum @mathhelpforum