The number of points on the curve y = 54x⁵ - 135x⁴ - 70x³ + 180x² + 210x at which the normal lines a

Q: Wait, why do we need to find the tangent slope instead of just using the normal slope directly?

That is a fantastic question! The derivative of a function, $\frac{dy}{dx}$, directly gives us the slope of the *tangent* line at any point. The problem gives us information about the *normal* line. Since the normal and tangent are perpendicular, their slopes are negative reciprocals of each other ($m_T \cdot m_N = -1$). We must convert the normal's slope to the tangent's slope to use the power of calculus to solve for $x$.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm for physics, but remember that this is a pure coordinate geometry and calculus problem! While physics problems often use energy conservation, here we are dealing with the geometric properties of curves and lines. We must stick to the tools of calculus—specifically, differentiation—to find the slope at specific points on the curve.

Q: Is it possible for the normal to be parallel to the line at the same point multiple times?

In this specific problem, we are looking for the *number of points* on the curve. When we solve the quartic equation $9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$, we are finding the $x$-coordinates of these points. Since all four roots are real and distinct, each $x$-value corresponds to a unique point on the curve. There is no overlap here; each root represents a distinct location where the condition is satisfied.

Question

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 :

Options:

(A) 2

(B) 3

(C) 4

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, today we embark on a journey to solve a problem that might look like a monster at first glance. We are presented with a fifth-degree polynomial, $y = 54x^5 - 135x^4 - 70x^3 + 180x^2 + 210x$, and a simple line, $x + 90y + 2 = 0$.

Our goal is to find the number of points on this curve where the normal line is parallel to our given line. Do not let the degree of the polynomial scare you; mathematics is often about finding the right perspective.

### The Geometric Bridge

First, let us understand the relationship between our lines. The given line is $x + 90y + 2 = 0$. If we rewrite this in the slope-intercept form, $y = mx + c$, we get:

$$y = -\frac{1}{90}x - \frac{2}{90}$$

Thus, the slope of our target line is $m_L = -\frac{1}{90}$. The problem states that the normal to the curve is parallel to this line. Since parallel lines share the same slope, the slope of our normal line must be $m_N = -\frac{1}{90}$.

Now, here is the crucial geometric insight: the tangent and the normal at any point on a curve are always perpendicular. This means their slopes must satisfy the condition $m_T \cdot m_N = -1$.

Substituting our value for $m_N$, we find that $m_T \cdot (-\frac{1}{90}) = -1$, which gives us a tangent slope of $m_T = 90$. This is the key that unlocks the entire problem.

### The Calculus Engine

Now that we know the tangent slope must be $90$, we turn to calculus. The derivative of our curve, $\frac{dy}{dx}$, represents the slope of the tangent at any point $x$.

Let us differentiate our polynomial term by term using the power rule:

$$\frac{dy}{dx} = 54(5x^4) - 135(4x^3) - 70(3x^2) + 180(2x) + 210$$

Simplifying this, we get:

$$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$$

We need to find the points where this slope equals $90$. So, we set up the equation:

$$270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$$

Subtracting $90$ from both sides, we arrive at:

$$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$$

### The Algebraic Battle

The coefficients are quite large, but look closely—they are all multiples of $30$. Let us divide the entire equation by $30$ to simplify our lives:

$$9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$$

Now, we need to find the roots of this quartic equation. We can use the hit and trial method.

Testing $x = 1$, we get $9(1)^4 - 18(1)^3 - 7(1)^2 + 12(1) + 4 = 9 - 18 - 7 + 12 + 4 = 0$. Success! $x = 1$ is a root.

Testing $x = 2$, we get $9(16) - 18(8) - 7(4) + 12(2) + 4 = 144 - 144 - 28 + 24 + 4 = 0$. Another success! $x = 2$ is also a root.

With two roots found, we can divide the quartic by $(x - 1)(x - 2) = x^2 - 3x + 2$. The resulting quadratic is $9x^2 + 9x + 2 = 0$.

Factoring this, we split the middle term to get $(3x + 1)(3x + 2) = 0$, yielding roots $x = -\frac{1}{3}$ and $x = -\frac{2}{3}$.

We have found four distinct real roots: $1, 2, -\frac{1}{3}, -\frac{2}{3}$. Each of these corresponds to a unique point on the curve. Therefore, there are exactly **4** points where the normal is parallel to the given line. You have conquered the polynomial!

Animated Solution for Mathematics - Differentiation: 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 :

Select Answer:

Visualized Solution

Analyze the Given Curve and Line

Slope of the Target Line

Slope of the Normal Line

Relate Normal to Tangent Slope

Differentiate the Curve Equation

Set up the Equation

Simplify the Quartic Equation

Factorize: Find First Root

Factorize: Find Second Root

Polynomial Division

Factorize the Quadratic

Identify the Number of Points

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Geometric Bridge

The Calculus Engine

The Algebraic Battle

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