Question

a. verify that the point is on the given curve. evaluate 4(x² + y²)² for x = 4 and y = 1.
4(x² + y²)² = 289xy²; (4,1)
b. determine an equation of the line tangent to the curve at the given point.

Explanation:

Step1: Evaluate left - hand side of the equation

Substitute \(x = 4\) and \(y=1\) into \(4(x^{2}+y^{2})^{2}\). First, calculate \(x^{2}+y^{2}\):
\(x^{2}+y^{2}=4^{2}+1^{2}=16 + 1=17\). Then \((x^{2}+y^{2})^{2}=17^{2}=289\), and \(4(x^{2}+y^{2})^{2}=4\times289 = 1156\).

Step2: Evaluate right - hand side of the equation

Substitute \(x = 4\) and \(y = 1\) into \(289xy^{2}\). We get \(289\times4\times1^{2}=289\times4=1156\).
Since the left - hand side \(4(x^{2}+y^{2})^{2}=1156\) and the right - hand side \(289xy^{2}=1156\) when \(x = 4\) and \(y = 1\), the point \((4,1)\) lies on the curve.

For part b, we first need to implicitly differentiate \(4(x^{2}+y^{2})^{2}=289xy^{2}\) with respect to \(x\).
Let \(u=x^{2}+y^{2}\), then the left - hand side is \(4u^{2}\). Using the chain rule, \(\frac{d}{dx}(4u^{2})=8u\frac{du}{dx}=8(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).
The right - hand side: Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = 289x\) and \(v=y^{2}\), we have \(\frac{d}{dx}(289xy^{2})=289y^{2}+289x\times2y\frac{dy}{dx}\).
So, \(8(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=289y^{2}+578xy\frac{dy}{dx}\).
Substitute \(x = 4\) and \(y = 1\):
First, \(x^{2}+y^{2}=17\).
The left - hand side is \(8\times17(2\times4+2\times1\times\frac{dy}{dx})=136(8 + 2\frac{dy}{dx})=1088+272\frac{dy}{dx}\).
The right - hand side is \(289\times1^{2}+578\times4\times1\times\frac{dy}{dx}=289 + 2312\frac{dy}{dx}\).
Then \(1088+272\frac{dy}{dx}=289+2312\frac{dy}{dx}\).
\(2312\frac{dy}{dx}-272\frac{dy}{dx}=1088 - 289\).
\(2040\frac{dy}{dx}=799\).
\(\frac{dy}{dx}=\frac{799}{2040}\).
The equation of the tangent line using the point - slope form \(y - y_{1}=m(x - x_{1})\) where \((x_{1},y_{1})=(4,1)\) and \(m=\frac{799}{2040}\) is \(y - 1=\frac{799}{2040}(x - 4)\), which simplifies to \(y=\frac{799}{2040}x-\frac{799}{510}+1=\frac{799}{2040}x-\frac{799 - 510}{510}=\frac{799}{2040}x-\frac{289}{510}\).

Answer:

a. The point \((4,1)\) lies on the curve since when \(x = 4\) and \(y = 1\), \(4(x^{2}+y^{2})^{2}=289xy^{2}=1156\).
b. The equation of the tangent line is \(y=\frac{799}{2040}x-\frac{289}{510}\)