Question

a. verify that the given point is on the curve. evaluate 48(x² + y²)² for x = 3 and y = 4. b. determine an equation of the line tangent to the curve at the given point. 48(x² + y²)² = 625xy²; (3,4)

Explanation:

Step1: Evaluate left - hand side for part a

Substitute \(x = 3\) and \(y = 4\) into \(48(x^{2}+y^{2})^{2}\). First, calculate \(x^{2}+y^{2}\):
\[x^{2}+y^{2}=3^{2}+4^{2}=9 + 16=25\]
Then, \((x^{2}+y^{2})^{2}=25^{2}=625\), and \(48(x^{2}+y^{2})^{2}=48\times625 = 30000\).

Step2: Evaluate right - hand side for part a

Substitute \(x = 3\) and \(y = 4\) into \(625xy^{2}\). We have \(xy^{2}=3\times4^{2}=3\times16 = 48\), and \(625xy^{2}=625\times48=30000\). 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 \(48(x^{2}+y^{2})^{2}=625xy^{2}\) implicitly

Let \(u=x^{2}+y^{2}\), then the left - hand side is \(48u^{2}\). Using the chain - rule, \(\frac{d}{dx}(48u^{2})=96u\frac{du}{dx}=96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})\).
The right - hand side: Using the product - rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = 625x\) and \(v = y^{2}\), we get \(\frac{d}{dx}(625xy^{2})=625y^{2}+1250xy\frac{dy}{dx}\).
So, \(96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=625y^{2}+1250xy\frac{dy}{dx}\).

Step4: Substitute \(x = 3\) and \(y = 4\) into the differentiated equation

First, \(x^{2}+y^{2}=25\). The left - hand side of the differentiated equation is \(96\times25\times(2\times3 + 2\times4\frac{dy}{dx})=2400\times(6 + 8\frac{dy}{dx})=14400+19200\frac{dy}{dx}\).
The right - hand side is \(625\times4^{2}+1250\times3\times4\frac{dy}{dx}=10000 + 15000\frac{dy}{dx}\).
Then, \(14400+19200\frac{dy}{dx}=10000 + 15000\frac{dy}{dx}\).
\[19200\frac{dy}{dx}-15000\frac{dy}{dx}=10000 - 14400\]
\[4200\frac{dy}{dx}=- 4400\]
\(\frac{dy}{dx}=-\frac{22}{21}\).

Step5: Find 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 \(m =-\frac{22}{21}\), we have \(y - 4=-\frac{22}{21}(x - 3)\).
\[21y-84=-22x + 66\]
\[22x+21y=150\]

Answer:

a. Since \(48(3^{2}+4^{2})^{2}=48\times25^{2}=48\times625 = 30000\) and \(625\times3\times4^{2}=625\times48 = 30000\), the point \((3,4)\) lies on the curve.
b. The equation of the tangent line to the curve at the point \((3,4)\) is \(22x + 21y=150\).