Question

a. verify that the given point lies on the curve. evaluate 48(x^{2}+y^{2})^{2} for x = 3 and y = 4. b. determine an equation of the line tangent to the curve at the given point. 48(x^{2}+y^{2})^{2}=625xy^{2},(3,4)

Explanation:

Step1: Substitute x and y values

Substitute \(x = 3\) and \(y=4\) into \(48(x^{2}+y^{2})^{2}\). First, calculate \(x^{2}+y^{2}\):
\[x^{2}+y^{2}=3^{2}+4^{2}=9 + 16=25\]

Step2: Calculate \((x^{2}+y^{2})^{2}\)

\((x^{2}+y^{2})^{2}=25^{2}=625\)

Step3: Calculate \(48(x^{2}+y^{2})^{2}\)

\[48\times625 = 30000\]
Now, calculate the right - hand side for \(625xy^{2}\) with \(x = 3\) and \(y = 4\):
\[625\times3\times4^{2}=625\times3\times16=625\times48 = 30000\]
Since the left - hand side \(48(x^{2}+y^{2})^{2}\) equals the right - hand side \(625xy^{2}\) when \(x = 3\) and \(y = 4\), the point \((3,4)\) lies on the curve.

For part b, we use implicit differentiation.
Let \(F(x,y)=48(x^{2}+y^{2})^{2}-625xy^{2}\)
First, find the partial derivatives:
\(\frac{\partial F}{\partial x}=48\times2(x^{2}+y^{2})\times2x-625y^{2}=192x(x^{2}+y^{2})-625y^{2}\)
\(\frac{\partial F}{\partial y}=48\times2(x^{2}+y^{2})\times2y-1250xy = 192y(x^{2}+y^{2})-1250xy\)
The slope of the tangent line \(m=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\)
At the point \((3,4)\):
\(\frac{\partial F}{\partial x}\big|_{(3,4)}=192\times3\times(3^{2}+4^{2})-625\times4^{2}=192\times3\times25-625\times16=14400 - 10000=4400\)
\(\frac{\partial F}{\partial y}\big|_{(3,4)}=192\times4\times(3^{2}+4^{2})-1250\times3\times4=192\times4\times25-15000=19200 - 15000 = 4200\)
\(m =-\frac{4400}{4200}=-\frac{22}{21}\)
Using the point - slope form \(y - y_{1}=m(x - x_{1})\) with \((x_{1},y_{1})=(3,4)\)
\(y - 4=-\frac{22}{21}(x - 3)\)
\(21y-84=-22x + 66\)
\(22x+21y=150\)

Answer:

a. The point \((3,4)\) lies on the curve since \(48(x^{2}+y^{2})^{2}=625xy^{2}\) holds for \(x = 3\) and \(y = 4\).
b. The equation of the tangent line is \(22x + 21y=150\)