Question

Question: 27-36 Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

\({y^{\bf{2}}}\left( {{y^{\bf{2}}} - {\bf{4}}} \right) = {x^{\bf{2}}}\left( {{x^{\bf{2}}} - {\bf{5}}} \right)\), \(\left( {{\bf{0}}, - {\bf{2}}} \right)\) (devil’s curve)

Short answer

The equation of the tangent to the curve is \(y = - 2\).

Step-by-step solution

Differentiate the equation of curve with respect to x

The equation of curve \({y^2}\left( {{y^2} - 4} \right) = {x^2}\left( {{x^2} - 5} \right)\) can be simplified as:

\(\begin{aligned}{y^2}\left( {{y^2} - 4} \right) &= {x^2}\left( {{x^2} - 5} \right)\\{y^4} - 4{y^2} &= {x^4} - 5{x^2}\end{aligned}\)

Differentiatethe equation \({y^4} - 4{y^2} = {x^4} - 5{x^2}\) on both sides.

\(\begin{aligned}4{y^3}\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) - 8y\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) &= 4{x^3} - 10x\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}}\left( {4{y^3} - 8y} \right) &= 4{x^3} - 10x\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{4{x^3} - 10x}}{{4{y^3} - 8y}}\end{aligned}\)

Find the value of \(y'\) at \(\left( {{\bf{0}}, - {\bf{2}}} \right)\)

Find the value of \(y'\left( {0, - 2} \right)\).

\(\begin{aligned}y'\left( {0, - 2} \right) &= \frac{{4{{\left( 0 \right)}^3} - 10\left( 0 \right)}}{{4{{\left( { - 2} \right)}^3} - 8\left( { - 2} \right)}}\\ &= 0\end{aligned}\)

So, the slope of the tangent to the curve at \(\left( {0, - 2} \right)\) is 0.

Step 3:Find the equation of the tangent at \(\left( {{\bf{0}}, - {\bf{2}}} \right)\)

The equation of the tangent at \(\left( {0, - 2} \right)\) can be determined as,

\(\begin{aligned}{c}y + 2 &= 0\left( {x - 0} \right)\\y + 2 &= 0\\y &= - 2\end{aligned}\)

Thus, the equation of tangent is \(y = - 2\).