maths

posted by .

The parabola y2 = 4ax, where a > 0, and the rectangular hyperbola xy = C2, where
C > 0, intersect at right angles. Show that the tangent and normal to either curve at the
point of intersection meet the x-axis at T and N where TN = 2pa, where p is an integer
to be determined.

  • maths -

    from y^2 = 4ax
    2y dy/dx = 4a
    dy/dx = 2a/y ---- >slope of tangent to parabola

    from xy = c^2
    xdy/dx + y = 0
    dy/dx = -y/x -----> slope of tangent of hyperbola

    but we are told that they intersect at right angles, so the tangents must be perpendicular, making
    2a/y = x/y
    x = 2a
    sub into xy=c^2
    y = c^2/(2a)

    so the intersect at (2a, c^2/(2a)
    (Now it gets messy)
    at parabola, slope = 2a/(c^2/2a) = 4a^2/c^2

    equation of tangent:
    y - c^2/(2a) = (4a^2/c^2)(x-2a)
    yc^2 - c^4/(2a) = 4a^2x - 8a^3
    at the x-axis, y = 0
    - c^4/(2a) = 4a^2x - 8a^3
    4a^2x = 8a^3 - c^4/(2a)
    x = 2a - c^4/(8a^3) -----> the x-intercept, call it T

    at the hyperbola, slope =
    slope = -y/x = (-c^2/(2a))/(2a) = -c^4/(4a^2)
    notice that this is the negative reciprocal of the slope at the parabola, so far so good!

    equation of tangent:
    (y - c^2/(2a)) = (-c^2/(4a^2)) (x - 2a)
    4a^2y - 2ac^2 = -c^2x + 2ac^2
    again at the x-intercept , let y = 0
    - 2ac^2 = -c^2x + 2ac^2
    c^2 x = 4ac^2
    x = 4a ------------> the other xintercept, call it N

    so TN = 4a - (2a - c^4/(8a^3)) = 2a + c^4/(8a^3)
    but TN = 2pa
    so
    2a + c^4/(8a^3) = 2pa
    p =1 + c^4/(16a^4)

    ????????? What do you think????

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Vectors

    I am really lost and confused! Consider the lines r = (1,-1,1) + t(3,2,1) and r = (-2,-3,0) + u(1,2,3). a) Find their point of intersection. b) Find a vector equation for the line perpendicular to both of the given lines that passes …
  2. maths

    Q.1 PQ RS are the two perpendicular chords of the rectangular hyperbola xy = c^2.If C is the center of the rectangular hyperbola,then the product of the slopes of CP,CQ,CR&CSis equal to______ Q.2If PN is the perpendicular from a point …
  3. calculus

    Determine the coordinates of the point of intersection of the two perpendicular lines that intersect on the y-axis and are both tangent to the parabola given below. y = 3x2
  4. math

    the tangent yo the curve y=x^2 +5x -2 @ the point (1,4)intersect the normal to the same curve @ the point (-3,-8) at the point P.Find the coordinates of point P.[ans: -1/3,-16/3] just give me some hint to calculate this solution.
  5. Calculus

    Draw a diagram to show that there are two tangent lines to the parabola y=x^2 that pass through the point (0,-4). Find the coordinates of the points where these tangent lines intersect the parabola. So far I have taken the derivative …
  6. Parabola Ques

    Find the point P on the parabola y^2 = 4ax such that area bounded by parabola, the X-axis and the tangent at P is equal to that of bounded by the parabola, the X-axis and the normal at P.
  7. Parabola ques

    the normal to a parabola y^2=4ax at the point t,where t is not 0 meets the curve again at the point t'.find t' in terms of t.determine a point on the x-axis where the tangent at t' meets the x-axis.
  8. math

    Draw a diagram to show that there are two tangent lines to the parabola y = x2 that pass through the point (0, −25). Find the coordinates of the points where these tangent lines intersect the parabola.
  9. maths

    A Variable Tangent To The Ellipse (x/a)^2 + (y/b)^2 =1 meets the parabola y^2=4ax at L and M. Find the locus of the midpoint of L M
  10. maths

    A variable tangent to the ellipse (x/a)^2 +(y/b)^2 meets the parabola y^2=4ax at L and M. Find the locus of the midpoint of LM.

More Similar Questions