Question

a. verify that the given point lies on the curve. 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) a. y = b. the point (3,4) does not lie on the curve.

Explanation:

Step1: Substitute x and y values

Substitute \(x = 3\) and \(y=4\) into the left - hand side of the equation \(48(x^{2}+y^{2})^{2}\).
\[

$$\begin{align*} x^{2}+y^{2}&=3^{2}+4^{2}\\ &=9 + 16\\ &=25 \end{align*}$$

\]
Then \((x^{2}+y^{2})^{2}=25^{2}=625\), and \(48(x^{2}+y^{2})^{2}=48\times625 = 30000\).

Step2: Substitute x and y values into the right - hand side

Substitute \(x = 3\) and \(y = 4\) into the right - hand side of the equation \(625xy^{2}\).
\[

$$\begin{align*} 625xy^{2}&=625\times3\times4^{2}\\ &=625\times3\times16\\ &=625\times48\\ &=30000 \end{align*}$$

\]
Since the left - hand side equals the right - hand side when \(x = 3\) and \(y = 4\), the point \((3,4)\) lies on the curve.

Step3: Differentiate the equation implicitly

Differentiate \(48(x^{2}+y^{2})^{2}=625xy^{2}\) with respect to \(x\).
Let \(u=x^{2}+y^{2}\), then \(\frac{d}{dx}[48(x^{2}+y^{2})^{2}]=48\times2(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).
The right - hand side: Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = 625x\) and \(v=y^{2}\), we have \(\frac{d}{dx}(625xy^{2})=625y^{2}+625x\times2y\frac{dy}{dx}\).
So \(96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=625y^{2}+1250xy\frac{dy}{dx}\).
Substitute \(x = 3\) and \(y = 4\) into the above equation:
First, \(x^{2}+y^{2}=3^{2}+4^{2}=25\).
\[

$$\begin{align*} 96\times25\times(2\times3 + 2\times4\frac{dy}{dx})&=625\times4^{2}+1250\times3\times4\frac{dy}{dx}\\ 96\times25\times(6 + 8\frac{dy}{dx})&=625\times16+15000\frac{dy}{dx}\\ 14400+19200\frac{dy}{dx}&=10000 + 15000\frac{dy}{dx}\\ 19200\frac{dy}{dx}-15000\frac{dy}{dx}&=10000 - 14400\\ 4200\frac{dy}{dx}&=- 4400\\ \frac{dy}{dx}&=-\frac{4400}{4200}=-\frac{22}{21} \end{align*}$$

\]
The equation of the tangent line using the point - slope form \(y - y_{1}=m(x - x_{1})\) where \((x_{1},y_{1})=(3,4)\) and \(m =-\frac{22}{21}\) is \(y - 4=-\frac{22}{21}(x - 3)\), which simplifies to \(21y-84=-22x + 66\), or \(22x+21y=150\).

Answer:

a. The point \((3,4)\) lies on the curve since when \(x = 3\) and \(y = 4\), \(48(x^{2}+y^{2})^{2}=625xy^{2}=30000\).
b. The equation of the tangent line to the curve at the point \((3,4)\) is \(22x + 21y=150\).