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Find the equation of the tangent of the parabola y^2= 4px

  1. Dec 13, 2009 #1
    hello everyone,
    :confused:i'm having can't seem to solve these two questions....

    1) Find the equation of the tangent of the parabola y^2= 4px, perpendicular to the lines 4Y- X + 3=0, and find the point of contact?

    2) Find the equation of the tangent to the parabola Y^2= 10x, at the extremities of the lactus rectum. On what line do these tangents intersect?

    If there is anyone out there who knows the answer...could u please explain or direct me to a suitable site which may explain the concepts.:cry:

    Thanks callas
  2. jcsd
  3. Dec 14, 2009 #2


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    Homework Helper

    Re: parabolas

    1) Can you find the gradient of that line? It says to find the tangent parallel to that line so when you take the derivative of the parabola, you're going to have to make it equal to the gradient of the line to find the value(s) of the points on the parabola that are have this gradient.

    2) Do you know how to find the equation of the latus rectum on a parabola? You'll have to again take the derivative of the parabola and then find what the gradients of the tangents are at the points where the latus rectum intersects the parabola. Since you'll know the it should be fairly simple to find the point of intersection between the tangent and the parabola, so you should be able to form an equation for each tangent. Now just find where these tangents intersect and notice the point's similarity to the focus and latus rectum.
  4. Dec 14, 2009 #3
    Re: parabolas

    thanks....but i can not use any calculus method....it must only be solved using analytic geometry......yes i could find the gradient....
  5. Dec 14, 2009 #4


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    Re: parabolas

    Any line perpendicular to 4y- x+ 3= 0 must have slope -4 and is of the form y= -4x+ a or x= -(1/4)y+ b. That line will cross to [itex]y^2= 4px[/itex] or [itex]x= (1/4p)y^2[/itex] where [itex]x= -(1/4)y+ b= y^2/4p[/itex] and will be tangent if and only if that equation has a double root. That is equivalent to the quadratic equation [itex]y^2+ py- 4pb= 0[/itex] and has a double root if and only if the discriminant, [itex]p^2- 4(-4pb)= p^2+ 16pb= 0[/itex]. Solve that for b.

    The "latus rectum" of this parabola is the vertical line segment, passing through the focus of the parabola with endpoints on the parabola. This particular parabola has focus at (2.5, 0) and its endpoints are where [itex]y^2= 10(2.5)[/itex] or (2.5, -5) and (2.5, 5).
    Any line passing through (2.5, -5) has the form x= m(y+5)+ 2.5. Again, use the fact that we must have a double root for [itex]x= (1/10)y^2= m(y+5)+ 2.5[/itex], which is equivalent to [itex]y^2- 10my+ 7= 0[/itex] to find m. Do the same with (2.5, 5).

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