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Step-by-step Solution
Step 1: Implicitly Differentiate the Given Curve
The curve is given by
4x^3 \;-\; 3xy^2 \;+\; 6x^2 \;-\; 5xy \;-\; 8y^2 \;+\; 9x \;+\; 14 \;=\; 0 .
To find the slope of the tangent at the point P(-2,\,3) , we differentiate both sides with respect to x , treating y as a function of x (i.e., y = y(x) ).
Step 2: Substitute the Coordinates into the Derivative
After differentiating, substitute x = -2 and y = 3 into the resulting expression to obtain y' , the slope of the tangent at P . From the given working:
y' \;=\; -\frac{9}{2}\,.
Step 3: Write the Equation of the Tangent Line
We know the slope of the tangent at P(-2, 3) is -\frac{9}{2} . The point-slope form for the line is
y - y_1 = m (x - x_1) . Thus,
y - 3 = -\frac{9}{2} (x + 2)\,.
\quad \Longrightarrow \quad 9x + 2y = -12\,.
Step 4: Write the Equation of the Normal Line
The slope of the normal is the negative reciprocal of the slope of the tangent. Hence, if m_{\text{tangent}} = -\frac{9}{2} , then
m_{\text{normal}} = \frac{2}{9}.
Using the point P(-2,\,3) again:
y - 3 = \frac{2}{9}(x + 2)
\quad \Longrightarrow \quad 9y - 2x = 31\,.
Step 5: Find the X-intercepts of the Tangent and Normal
To find where each line meets the x -axis, set y = 0 in each equation.
Tangent ( 9x + 2y = -12 ):
Set y = 0, then 9x = -12 \;\Longrightarrow\; x = -\frac{4}{3}.
So the tangent meets the x -axis at (-\tfrac{4}{3},\,0).
Normal ( 9y - 2x = 31 ):
Set y = 0, then -2x = 31 \;\Longrightarrow\; x = -\frac{31}{2}.
So the normal meets the x -axis at (-\tfrac{31}{2},\,0).
Step 6: Calculate the Enclosed Area
The triangle formed by the x -axis, the tangent, and the normal has vertices at
P \;(-2,\,3),\; T \;\bigl(-\tfrac{4}{3},\,0\bigr),\; N \;\bigl(-\tfrac{31}{2},\,0\bigr).
The base of this triangle lies along the x -axis, between
x = -\tfrac{31}{2} and x = -\tfrac{4}{3}.
Base length = \left|\,-\frac{31}{2} \;-\; \left(-\frac{4}{3}\right)\right|
= \left|\,-\frac{31}{2} + \frac{4}{3}\right|
= \left|\,-\frac{93}{6} + \frac{8}{6}\right|
= \frac{85}{6}.
The height of this triangle is the y -coordinate of the point P , which is 3 , because P projects vertically onto the x -axis to enclose the triangle.
Thus, the area A is
A = \frac{1}{2} \times \bigl(\text{base}\bigr) \times \bigl(\text{height}\bigr)
= \frac{1}{2} \times \frac{85}{6} \times 3
= \frac{85}{4}\,.
Step 7: Compute 8A
We need 8A , so
8A = 8 \times \frac{85}{4} = 2 \times 85 = 170\,.
Final Answer
\displaystyle 8A = 170\,.
