Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Slopes of the tangent line?

Determine the exact coordinates for the four points that are the x andyintercepts of the graph of the equation 10x^2+ 5y^2+ 4xy= 110. Find the slopes of the tangent lines for each of the points.

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  • david
    Lv 7
    1 month ago
    Best Answer

    10x^2+ 5y^2+ 4xy= 110

    for the x-intercept, y = 0

    10x^2 = 110

    x = +/- sqrt(11)

    for the y-intercept. x = 0

    5y^2 = 110.

    y = +/- sqrt(22)

    ======================================================

    10x^2+ 5y^2+ 4xy = 110

    20dx + 10ydy + 4(xdy + ydx) = 0

    20dx + 4ydx + 10ydy + 4xdy = 0.

    (20 + 4y)dx = -(10y + 4x)dy

    dy/dx = -(20 + 4y)/(10y + 4x)

    at (0,sqrt(22)) ... m = -(20sqrt(22) + 4)/(10sqrt(22))

    . . . . m = -2(sqrt(22) + 4)/(sqrt(22))

    ==== this isn't fun anymore, just a bunch of algebra involving sq roots, === so since it is your assignment, you can finish.

    • ted s
      Lv 7
      1 month agoReport

      ' david ' missed an x in the 2nd line...20 x dx

  • Ash
    Lv 7
    1 month ago

    10x²+ 5y²+ 4xy= 110

    To find x-intercepts, put y=0

    10x²+ 5(0)²+ 4x(0)= 110

    10x² = 110

    x² = 11

    x = ±√11

    x-intercepts are (√11,0) and (-√11,0)

    To find y-intercepts, put x=0

    10(0)²+ 5y²+ 4(0)y= 110

    5y² = 110

    y² = 22

    y = ±√22

    y-intercepts are (0,√22) and (0,-√22)

    To find slope of tangent, find derivative of 10x²+ 5y²+ 4xy= 110

    20x+ 10y dy/dx+ 4(y + x dy/dx) = 0

    20x+ 10y dy/dx+ 4y + 4x dy/dx = 0

    dy/dx(10y+4x) = -20x-4y

    dy/dx = -4(5x+y)/2(5y+2x)

    dy/dx = -2(5x+y)/(5y+2x)

    At (√11,0), slope dy/dx = -2(5√11+0)/(5*0+2√11) = -(10√11)/(2√11) = -5

    Using the formula y-y₁=m(x-x₁)

    equation of tangent is y-0 = -5(x-√11)

    y = -5x + 5√11

    At (-√11,0), slope dy/dx = -2(5*-√11+0)/(5*0+2*-√11) = (10√11)/(-2√11) = -5

    equation of tangent is y-0 = -5(x-(-√11))

    y = -5x - 5√11

    At (0,√22), slope dy/dx = -2(5*0+√22)/(5*√22 +2*0) = -(2√22)/(5√22) = -2/5

    equation of tangent is y-√22 = (-2/5)(x-0)

    y = -(2/5)x + √22

    At (0,-√22), slope dy/dx = -2(5*0-√22)/(5*-√22 +2*0) = (2√22)/(-5√22) = -2/5

    equation of tangent is y-(-√22) = (-2/5)(x-0)

    y = -(2/5)x - √22

    If you want to see how the function and tangents look then click on below link

    https://www.desmos.com/calculator/j0tzfh3apy

  • 1 month ago

    When x = 0, you have 5y^2 = 110, so

    y = +/- sqrt(22), around 4.7.

    When y = 0, you have 10x^2 = 110, so

    x = +/- sqrt(11), around 3.3.

    Implicit differentiation:

    20x + 25y dy/dx + 4y + 4x dy/dx = 0.

    When x = 0, dy/dx = -4y/25y = -0.16 at both of the y-intercepts.

    When y = 0, dy/dx = -20x/4x = -5 at both of the x-intercepts.

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