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Locus of point of intersection of tangents

  1. Sep 13, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Locus of the point of intersection of tangents to the parabolas y[itex]^{2}[/itex]=4(x+1) and y[itex]^{2}[/itex]=8(x+2) which are at right angles, is


    2. Relevant equations
    Equation of tangent for first parabola
    t[itex]_{1}[/itex]y=x+1+at[itex]_{1}[/itex][itex]^{2}[/itex]
    Equation of tangent for second parabola
    t[itex]_{2}[/itex]y=x+2+bt[itex]_{2}[/itex][itex]^{2}[/itex]

    3. The attempt at a solution
    Let us assume that the point of intersection of tangents is (h,k)
    Since it lies on the tangent
    ∴kt[itex]_{1}[/itex]=h+1+at[itex]_{1}[/itex][itex]^{2}[/itex]
    [itex]\Rightarrow[/itex]at[itex]_{1}[/itex][itex]^{2}[/itex]-kt[itex]_{1}[/itex]+(h+1)=0
    t[itex]_{1}[/itex]t[itex]_{2}[/itex]=h+1 (product of roots)
    Also t[itex]_{1}[/itex]t[itex]_{2}[/itex]=-1 (Since they are at right angles)
    ∴required locus = x+2=0

    I can't understand where I'm wrong. Is the equation of tangent incorrect? Please Help.
     
  2. jcsd
  3. Sep 13, 2012 #2

    ehild

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    I can not follow your derivation. What have you got for a and b to make the equations tangent to the parabolas? The product of roots in the equation in red is (h+1)/a. t2 is not the other root of the equation for the first tangent line. Use the equations of both tangent lines to get the intersection.

    I also got x=const. for the locus, but different from yours.

    ehild
     
    Last edited: Sep 13, 2012
  4. Sep 14, 2012 #3

    utkarshakash

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    Hey actually i arrived at the wrong answer because I only considered the first equation but not second. By solving both I get x+2=0 and x+4=0. Adding these two yields 2x+6=0 or x+3=0 which is the required locus.

    P.S.- I wrote product of roots as h+1 because a=1. so (h+1)/a=(h+1)/1=(h+1) :wink:
     
  5. Sep 14, 2012 #4

    ehild

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    x+2=0 and x+4=0 contradict to each other, so that is not the right way to go, but the end result is correct. I got the same .

    But you kept that a=1 top secret. It took me some time to find out what your equations mean.

    ehild
     
  6. Sep 15, 2012 #5

    ehild

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    As it is a nice problem, and the OP has solved it, I write out the steps in a more ordered form.

    We have two parabolas

    (1)[tex]y^2=4(x+1)[/tex]
    (2)[tex]y^2=8(x+2)[/tex]

    The tangent lines are
    [tex]x=ty+q[/tex].
    Find q in terms of t so the line is tangent to the parabola, that is, the line and the parabola have a single common point. Substitute for x into the equation of parabola(1):
    (1)
    [tex]y^2-4t_1y-4(q_1+1)=0[/tex]
    The quadratic equation has a single solution, so the discriminant is zero--> [tex]D=16t_1^2+16(q_1+1)=0\rightarrow q_1=-(t_1^2+1)[/tex]
    (2)Substitute the equation of the tangent line for x in the equation of the second parabola: [tex]y^2-8t_21y-8(q_1+2)=0[/tex]
    The discriminant is [tex]D=64t_2^2+32(q_2+2)=0\rightarrow q_2=-2(t_1^2+1)[/tex]

    The tangent lines are perpendicular so [tex]t_1t_2=-1[/tex]
    Let t1=t, t2=-1/t. With that notation:
    (1)[tex]x=ty-(t^2+1) \rightarrow y=x/t+(t+1/t)[/tex]
    (2)[tex]x=-y/t-2(-1/t^2+1) \rightarrow y=-tx-2(t+1/t)[/tex]
    Subtract the expressions for y to get the abscissa of the locus.

    ehild
     
  7. Sep 15, 2012 #6

    utkarshakash

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    Thanks for helping.
     
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