The number of points on the curve \(y=54 x^5-135 x^4-70 x^3+180 x^2+210 x\) at which the normal lines are parallel \(to x+90 y+2=0\) is
The number of points on the curve \(y=54 x^5-135 x^4-70 x^3+180 x^2+210 x\) at which the normal lines are parallel \(to x+90 y+2=0\) is
- 2
- 4
- 0
- 3
The Correct Option is B
Approach Solution - 1
Equation of the Curve
The equation of the curve is:
\[ y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x. \]
Step 1: Slope of the Normal Line
The normal line is parallel to the line \(x + 90y + 2 = 0\), which has a slope:
\[ m = -\frac{1}{90}. \]
The slope of the normal line is also \(m_N = -\frac{1}{90}\). The normal slope is related to the derivative of the curve by:
\[ m_N = -\frac{1}{\frac{dy}{dx}}. \]
Equating the slopes:
\[ -\frac{1}{90} = -\frac{1}{\frac{dy}{dx}}. \]
This simplifies to:
\[ \frac{dy}{dx} = 90. \]
Step 2: Compute \(\frac{dy}{dx}\)
The derivative of the curve is:
\[ \frac{dy}{dx} = \frac{d}{dx} \left(54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x\right). \]
Differentiate each term:
\[ \frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210. \]
We are given that \(\frac{dy}{dx} = 90\), so substitute and simplify:
\[ 270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90. \]
Subtract 90 from both sides:
\[ 270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0. \]
Step 3: Solve the Equation
The equation:
\[ 270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0 \]
has 4 real roots, corresponding to 4 points on the curve.
Conclusion
The number of points where the normal lines are parallel to \(x + 90y + 2 = 0\) is:
\[ \boxed{4}. \]
Approach Solution -2
Normal of line is parallel to line x+90y+2=0
\(m_N=−\frac{1}{90}\)
\(−(\frac{dx}{dy})_{(x_1y_1)}=−\frac{1}{90}⇒(\frac{dy}{dx})_{(x_1y_1)}=90\)
Now,
\(\frac{dy}{dx}=270x^4−540x^3−210x^2+360x+210=90\)
\(⇒x=1,2,\frac{−2}{3},\frac{−1}{3}\)
so, the correct option is(B): 4 normals
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