Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Calculate the slope of the normal line to the indicated curve at the given point x^2−y=54 at (−9,27)?

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  • 2 months ago
    Favourite answer

    x^2−y=54 at (−9,27)

    that is 

    y = x² – 54

    first check if that point is on the curve

    27 = (–9)² – 54 = 81–54

    ok

    that is y = x² – 54y' = 2xy'(–9) = –18that is the slope of the tangent.perpendicular, slope would be +1/18so equation isy = (1/18)x + kuse point to find k27 = (1/18)(–9) + kk = 27 + (1/2) = 55/2equation is y = (1/18)x + (55/2)

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  • 2 months ago

    x^2 - y = 54

    -y = -x^2 + 54

    y = x^2 - 54

    dy/dx = 2x

    dy/dx(at x = -9) = 2(-9) = -18

    slope of the normal line is -1/(-18) = 1/18

  • Philip
    Lv 6
    2 months ago

    Given curve is y = x^2-54 with y' = 2x = slope of tangent at (x,y). Slope of

    normal at (x,y) is -1/(2x). Slope of normal line at (-9,27) is -1/(2*(-9)) = (1/18).

  • Ian H
    Lv 7
    2 months ago

    y = x^2 – 54, point P(−9 ,27)

    tangent slope dy/dx = 2x = -18, normal slope 1/18

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  • 2 months ago

    y = ^2 - 54 

    Differentiate 

    dy/dx = 2x   ( The slope of the tangent t any point 'x'. 

    When x = -9 

    dy/dx = -18

    Hence slope of the normal is the negative reciprocal  which is '1/18'. 

    The normal line is

    y -27 = (1/18)(x - - 9) 

    y - 27 = x/18 + 9/18 

    y = x / 18 + 1/2 + 27 

    y = x / 18 + 55/2   

  • Pope
    Lv 7
    2 months ago

    x² - y = 54

    x² = y + 54

    The curve is a parabola having a vertical axis.

    Vertex: (0, -54)

    Latus rectum: 1

    Axis: x = 0

    Point P(-9, 27) is on the curve.

    The abscissa of P is greater than half the latus rectum.

    Its projection image on the axis is Q(0, 27).

    Translate Q away from the vertex by half the latus rectum, to R(0, 27.5).

    The normal at P is line PR.

    slope = (27.5 - 27) / (0 + 9) = 1/18

  • 2 months ago

    We can rearrange for y explicitly as

    y =  x² - 4

    Differentiating gives

    dy/dx =  2x

    Recall the derivative gives the slope of the line tangent to y at any x. Thus, at x = -9, the slope of the tangent is -18.

    Recall that line M perpendicular to a line with slope m will have slope -1/m.

    Thus, in our case the normal line has slope 1/18.

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