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)?

Relevance

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) • 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

• 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).

• y = x^2 – 54, point P(−9 ,27)

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

• 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

• 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

• 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.