The number of points on the curve y=54x^5-135 | vedclass

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(C) The equation of the given line is $x+90y+2=0$,which can be written as $y=-\frac{1}{90}x-\frac{2}{90}$.The slope of this line is $m=-\frac{1}{90}$.

Vedclass · Accepted Answer

$4$

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(C) The equation of the given line is $x+90y+2=0$,which can be written as $y=-\frac{1}{90}x-\frac{2}{90}$.The slope of this line is $m=-\frac{1}{90}$.Since the normal line is parallel to this line,the slope of the normal $(m_N)$ must be equal to $-\frac{1}{90}$.The slope of the tangent $(m_T)$ is the negative reciprocal of the slope of the normal,so $m_T = -\frac{1}{m_N} = -\frac{1}{-1/90} = 90$.We find the derivative of the curve $y=54x^5-135x^4-70x^3+180x^2+210x$:$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$.Setting the derivative equal to the slope of the tangent $(90)$:$270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$Dividing by $30$:$9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$.Testing for roots,we find the roots are $x=1, x=2, x=-\frac{2}{3}, x=-\frac{1}{3}$.Since there are $4$ distinct values of $x$,there are $4$ such points on the curve.

The number of points on the curve $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$ is:

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