Find an equation of the tangent line to the curve $120 \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2312 \left( x ^ { 2 } - y ^ { 2 } \right)$ at the point $( 4,1 )$.
Select the correct answer.
A) $y = - 1.11 x + 17$
B) $y = - 1.11 x + 3.43$
C) $y = 1.11 x + 5.43$
D) $y = - 1.11 x + 5.43$
E) None of these

Question

Quizplus · Accepted Answer

First, we need to find the slope of the tangent line at the point $(4,1)$. 
We can do this by taking the derivative of the equation with respect to $x$:

\begin{align*}
\frac{d}{dx}\left[ 120 \left( x ^ { 2 } + y ^ { 2 } ight) ^ { 2 } ight] &= \frac{d}{dx}\left[ 2312 \left( x ^ { 2 } - y ^ { 2 } ight) ight]\
240 \left( x ^ { 2 } + y ^ { 2 } ight) \cdot 2x + 240 \left( x ^ { 2 } + y ^ { 2 } ight) \cdot 2y \cdot \frac{dy}{dx} &= 4624x-4624y \frac{dy}{dx}\
480x^3 + 480xy^2 + 4624y \frac{dy}{dx} &= 4624x - 480xy^2 - 4624y \frac{dy}{dx}\
9248y \frac{dy}{dx} &= 4624x - 960x^3\
\frac{dy}{dx} &= \frac{4624x - 960x^3}{9248y}\
\end{align*}

Now, we substitute $(4,1)$ into this expression to find the slope:

\begin{align*}
\left. \frac{dy}{dx} ight| _{ (4,1) } &= \frac{4624(4) - 960(4)^3}{9248(1)}\
&= -\frac{11}{10} 
\end{align*}

Therefore, the equation of the tangent line passing through $(4,1)$ can be expressed as:

\begin{align*}
y - 1 &= -\frac{11}{10}(x-4) \
y  &= -\frac{11}{10}(x-4)+1 \
y  &= -\frac{11}{10}x +\frac{53}{5} \
y  &= -1.1x+10.6 \
\end{align*}

Therefore, the correct choice is D.

Find an Equation of the Tangent Line to the Curve 120(x2+y2)2=2312(x2−y2)120 \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2312 \left( x ^ { 2 } - y ^ { 2 } \right)120(x2+y2)2=2312(x2−y2)

Find an Equation of the Tangent Line to the Curve $120 \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2312 \left( x ^ { 2 } - y ^ { 2 } \right)$