Find the equation of the line normal to the curve defined by the equation $x ^ { 3 } y ^ { 4 } - 5 = x ^ { 3 } - x ^ { 2 } + y$ at the point (2, $- 1$ ).
A) $33 x + 4 y - 62 = 0$
B) $33 x - 4 y - 62 = 0$
C) $33 x + 4 y + 62 = 0$

D) $33 x - 4 y + 62 = 0$
E) $4 x - 33 y - 41 = 0$
F) $4 x + 33 y - 41 = 0$
G) $4 x - 33 y + 41 = 0$
H) $4 x + 33 y + 41 = 0$

Question

Find the equation of the line normal to the curve defined by the equation $x ^ { 3 } y ^ { 4 } - 5 = x ^ { 3 } - x ^ { 2 } + y$ at the point (2, $- 1$ ).
A) $33 x + 4 y - 62 = 0$
B) $33 x - 4 y - 62 = 0$
C) $33 x + 4 y + 62 = 0$

D) $33 x - 4 y + 62 = 0$
E) $4 x - 33 y - 41 = 0$
F) $4 x + 33 y - 41 = 0$
G) $4 x - 33 y + 41 = 0$
H) $4 x + 33 y + 41 = 0$

Quizplus · Accepted Answer

To find the equation of the line normal to the curve at a given point, we first need to find the slope of the tangent to the curve at that point, and then find the slope of the normal line, which is the negative reciprocal of the tangent's slope. The given curve is $x^3y^4 - 5 = x^3 - x^2 + y$.1. Differentiate both sides with respect to $x$ to find the slope of the tangent line ($dy/dx$):$$3x^2y^4 + 4x^3y^3(dy/dx) = 3x^2 - 2x + (dy/dx)$$2. Solve for $dy/dx$ at the point (2, -1):Substitute $x = 2$ and $y = -1$ into the differentiated equation to find $dy/dx$.3. Find the slope of the normal line, which is the negative reciprocal of $dy/dx$.4. Use the point-slope form of a line equation, $y - y_1 = m(x - x_1)$, where $m$ is the slope of the normal line, and $(x_1, y_1)$ is the point (2, -1), to find the equation of the normal line.Following these steps, the correct equation of the normal line is found to be $33x + 4y - 62 = 0$, which corresponds to choice A.

Find the Equation of the Line Normal to the Curve x3y4−5=x3−x2+yx ^ { 3 } y ^ { 4 } - 5 = x ^ { 3 } - x ^ { 2 } + yx3y4−5=x3−x2+y

Find the Equation of the Line Normal to the Curve $x ^ { 3 } y ^ { 4 } - 5 = x ^ { 3 } - x ^ { 2 } + y$