Question

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 

• A. 2
• B. 4
• C. 0
• D. 3

Answer 1

The Correct Option is B

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:

\[ \boxed{4}. \]

Answer 2

Normal of line is parallel to line x+90y+2=0
\(m_N​=−\frac{1}{90}​\)
\(−(\frac{dx}{dy}​)_{(x_1​y_1​)​}=−\frac{1}{90}⇒(\frac{dy}{dx}​)_{(x_1​y_1​)​}=90\)
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