Results 1 to 3 of 3

Math Help - Addition of Intercepts for curve

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    2

    Addition of Intercepts for curve

    Hi

    I need help with this question:
    a curve
    square root(x) + Square root (y) = 1(eqn 1), for x and y bigger and equal too zero

    Show that the sum of x any y intercepts of the tangent line to any point (a,b) on the curve (a,b bigger than 0) is equal to one

    This is what i have done so far

    Used implicit differentiation to find dy/dx of the eqn 1
    It gives
    dy/dx=-sqrt(y)/sqrt(x)

    from there used y=mx+c

    So m=-sqrt(y)/sqrt(x)
    an then i subbed (a,b) into dy/dx to find m

    so
    that is y=(-sqrt(b)/sqrt(a))x+ c

    then i subbed a(a,b) into y=(-sqrt(b)/sqrt(a))x+ c to find c
    i found c= b+sqrt(ab)

    from there i tried to find the x-intercept(used y=0) an y-intercept(used x=0)

    so i found that x-int=sqrt(ab)-a
    and y-int=b+sqrt(ab)

    from there i've tried to relate it to original equation somehow but cannot relate it so that the sum of the intercepts equals 1

    Any help would be appreciated, thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,307
    Thanks
    1284
    Quote Originally Posted by Kevinsalborn View Post
    Hi

    I need help with this question:
    a curve
    square root(x) + Square root (y) = 1(eqn 1), for x and y bigger and equal too zero

    Show that the sum of x any y intercepts of the tangent line to any point (a,b) on the curve (a,b bigger than 0) is equal to one

    This is what i have done so far

    Used implicit differentiation to find dy/dx of the eqn 1
    It gives
    dy/dx=-sqrt(y)/sqrt(x)

    from there used y=mx+c

    So m=-sqrt(y)/sqrt(x)
    an then i subbed (a,b) into dy/dx to find m

    so
    that is y=(-sqrt(b)/sqrt(a))x+ c
    A simpler form is y= m(x- x_0)+ y_0 giving y= \frac{-\sqrt{b}}{\sqrt{a}}(x- a)+ b. That immediately gives y= \frac{\sqrt{a}}{\sqrt{b}}x+ \sqrt{ab}+ b as you have.

    then i subbed a(a,b) into y=(-sqrt(b)/sqrt(a))x+ c to find c
    i found c= b+sqrt(ab)


    from there i tried to find the x-intercept(used y=0) an y-intercept(used x=0)

    so i found that x-int=sqrt(ab)-a
    and y-int=b+sqrt(ab)

    from there i've tried to relate it to original equation somehow but cannot relate it so that the sum of the intercepts equals 1

    Any help would be appreciated, thank you
    Very good!
    With x= 0, y= -\frac{\sqrt{b}}{\sqrt{a}}(-a)+ b=  b+ \sqrt{ab} and with y= 0, 0= -\frac{\sqrt{b}}{\sqrt{a}}x+ b+ \sqrt{ab} so \frac{\sqrt{b}}{\sqrt{a}}x= b+ \sqrt{ab}, x= \frac{\sqrt{a}}{\sqrt{b}}b+ \frac{\sqrt{a}}{\sqrt{b}}\sqrt{ab}= a+ \sqrt{ab}. That is not exactly what you got- notice the "+ a" in the y-intercept.

    x-int+ y-int= b+ \sqrt{ab}+ a+ \sqrt{b}= b+ 2\sqrt{a}\sqrt{b}+ a and you ought to be able to recognize that as a "perfect square": (\sqrt{a}+ \sqrt{b})^2!

    Now, use the fact that (a, b) is on the given curve so \sqrt{a}+ \sqrt{b}= 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    2
    I see it now, it was a perfect square, thank you so much for that
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: February 23rd 2011, 03:46 AM
  2. Y intercepts
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 13th 2010, 09:58 AM
  3. X & Y Intercepts
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 30th 2009, 11:21 PM
  4. Determining x-intercepts of a curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 16th 2009, 10:06 PM
  5. how do i find both x intercepts of a curve?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2008, 11:23 AM

Search Tags


/mathhelpforum @mathhelpforum