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) evaluate 625xy^2 for x = 3 and y = 4.

Explanation:

Step1: Evaluate the expression for part - left

Substitute \(x = 3\) and \(y = 4\) into \(625xy^{2}\).
\[625\times3\times4^{2}\]

Step2: Calculate the power

First, calculate \(4^{2}=16\).
\[625\times3\times16\]

Step3: Multiply the numbers

\[625\times3 = 1875\], then \(1875\times16=30000\).

Step4: Verify point for part - right (a)

Substitute \(x = 3\) and \(y = 4\) into the left - hand side of the equation \(48(x^{2}+y^{2})^{2}\).
First, calculate \(x^{2}+y^{2}\): \(3^{2}+4^{2}=9 + 16=25\).
Then \((x^{2}+y^{2})^{2}=25^{2}=625\), and \(48(x^{2}+y^{2})^{2}=48\times625 = 30000\).
Substitute \(x = 3\) and \(y = 4\) into the right - hand side of the equation \(625xy^{2}\), we get \(625\times3\times4^{2}=30000\). Since the left - hand side equals the right - hand side, the point \((3,4)\) lies on the curve.

Step5: Find the derivative for part - right (b)

Differentiate \(48(x^{2}+y^{2})^{2}=625xy^{2}\) implicitly with respect to \(x\).
Using the chain rule and product rule:
Let \(u=x^{2}+y^{2}\), then the left - hand side is \(48\times2u\times(2x + 2y\frac{dy}{dx})=96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).
The right - hand side: Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\) where \(u = 625x\) and \(v = y^{2}\), we have \(625y^{2}+1250xy\frac{dy}{dx}\).
\[96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=625y^{2}+1250xy\frac{dy}{dx}\]
Substitute \(x = 3\) and \(y = 4\):
First, \(x^{2}+y^{2}=25\).
\[96\times25\times(2\times3 + 2\times4\frac{dy}{dx})=625\times4^{2}+1250\times3\times4\frac{dy}{dx}\]
\[2400\times(6 + 8\frac{dy}{dx})=10000+15000\frac{dy}{dx}\]
\[14400+19200\frac{dy}{dx}=10000 + 15000\frac{dy}{dx}\]
\[19200\frac{dy}{dx}-15000\frac{dy}{dx}=10000 - 14400\]
\[4200\frac{dy}{dx}=- 4400\]
\(\frac{dy}{dx}=-\frac{22}{21}\)

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

Answer:

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