Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ - 3xy² + 6x²

Q: Wait, why can't I just solve the curve equation for $y$ and then differentiate?

That is a common trap! The equation $4x^3 - 3xy^2 + 6x^2 - 5xy - 8y^2 + 9x + 14 = 0$ is an implicit function. Solving for $y$ would involve the quadratic formula and result in a messy expression with square roots, making differentiation a nightmare. Implicit differentiation is the elegant, professional way to handle this.

Q: Do I really need to find the full equation of the normal line?

Absolutely! The triangle is defined by three vertices: the point of tangency $P$, the x-intercept of the tangent $T$, and the x-intercept of the normal $N$. Without the normal's equation, you cannot find point $N$, and thus you cannot determine the base of the triangle.

Q: Why is the base of the triangle $|x_T - x_N|$?

Since both the tangent and the normal intersect the x-axis, the segment connecting these two points of intersection lies entirely on the x-axis. The length of this segment is simply the distance between the two x-coordinates, which is the absolute difference $|x_T - x_N|$.

Question

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $4x^3 - 3xy^2 + 6x^2 - 5xy - 8y^2 + 9x + 14 = 0$ at the point $(-2, 3)$ be $A$. Then $8A$ is equal to ________

Sigma Percentile · Accepted Answer

### The Geometry of Curves: A Tangent and Normal Odyssey

Imagine standing on a complex, winding path defined by the equation $4x^3 - 3xy^2 + 6x^2 - 5xy - 8y^2 + 9x + 14 = 0$. You are standing at a specific point $P(-2, 3)$.

Your mission is to construct the tangent and normal lines at this point and calculate the area of the triangle they form with the x-axis. This is a dance of geometry and calculus.

### Phase 1: The Calculus of Slopes

To find the tangent, we need its slope. Since our curve is implicit, we use implicit differentiation.

We differentiate term by term with respect to $x$, remembering that every time we differentiate a $y$ term, the chain rule demands we multiply by $y'$. Applying this to our curve, we get:

$$12x^2 - 3(y^2 + 2xyy') + 12x - 5(y + xy') - 16yy' + 9 = 0$$

Instead of isolating $y'$ algebraically, we substitute $x = -2$ and $y = 3$ directly into this equation. This transforms our complex derivative into a simple linear equation:

$$48 - 3(9 - 12y') - 24 - 15 + 10y' - 48y' + 9 = 0$$

Solving this, we find the slope of the tangent $m_T = -\frac{9}{2}$.

### Phase 2: The Geometry of Lines

With the tangent slope $m_T = -\frac{9}{2}$, the normal slope $m_N$ must be the negative reciprocal, $m_N = \frac{2}{9}$, because the normal is perpendicular to the tangent.

Now, we use the point-slope form $y - y_1 = m(x - x_1)$ to find the equations of these lines.

For the tangent, $y - 3 = -\frac{9}{2}(x + 2)$. Setting $y = 0$ reveals the x-intercept $T$ at $x = -\frac{4}{3}$.

For the normal, $y - 3 = \frac{2}{9}(x + 2)$. Setting $y = 0$ gives the x-intercept $N$ at $x = -\frac{31}{2}$.

### Phase 3: The Final Area

We now have our three vertices: $P(-2, 3)$, $T(-\frac{4}{3}, 0)$, and $N(-\frac{31}{2}, 0)$.

The base of our triangle lies on the x-axis, with length:

$$|x_T - x_N| = |-\frac{4}{3} - (-\frac{31}{2})| = \frac{85}{6}$$

The height is the vertical distance from the x-axis to $P$, which is simply the y-coordinate of $P$, $h = 3$.

The area $A$ is calculated as:

$$A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{85}{6} 	imes 3 = \frac{85}{4}$$

Finally, the question asks for $8A$. Multiplying our result by 8, we get:

$$8 	imes \frac{85}{4} = **170**$$

We have successfully navigated the curve, mastered the calculus, and uncovered the geometric truth hidden within the algebra.

Animated Solution for Mathematics - Differentiation: Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $4 x^{3} - 3 x y^{2} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$ at the point $(- 2, 3)$ be $A$ . Then $8 A$ is equal to ________

Enter Numerical Value:

Visualized Solution

Visualizing the Curve and Point $P$

Implicit Differentiation

Applying the Product Rule

Substituting Point $P (- 2, 3)$

Solving for Slope $y^{'}$

Slopes of Tangent and Normal

Tangent's X-intercept

Normal's X-intercept

Visualizing the Triangle Area

Calculating the Base of the Triangle

Finding the Area $A$

Final Answer: $8 A$

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Curves

Phase 1

Phase 2

Phase 3

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Curve and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Implicit Differentiation

Applying the Product Rule

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−2,3)

Solving for Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′