Question

1. verify that the point is on the given curve. evaluate 4(x² + y²)² for x = 1 and y = 2. 2. determine an equation of the line tangent to the curve at the given point. (x² + y²)² = 25xy²,(1,2)

Explanation:

Step1: Substitute x and y values

Substitute \(x = 1\) and \(y=2\) into \(4(x^{2}+y^{2})^{2}\). First, calculate \(x^{2}+y^{2}\):

$$x^{2}+y^{2}=1^{2}+2^{2}=1 + 4=5$$

Step2: Calculate the square - squared value

Then calculate \((x^{2}+y^{2})^{2}\) and multiply by 4. Since \((x^{2}+y^{2}) = 5\), then \((x^{2}+y^{2})^{2}=5^{2}=25\), and \(4(x^{2}+y^{2})^{2}=4\times25 = 100\).
Next, for the right - hand side of the equation \(25xy^{2}\), substitute \(x = 1\) and \(y = 2\):

$$25xy^{2}=25\times1\times2^{2}=25\times4=100$$

Since the left - hand side \(4(x^{2}+y^{2})^{2}=100\) and the right - hand side \(25xy^{2}=100\) when \(x = 1\) and \(y = 2\), the point \((1,2)\) is on the curve.

Now, to find the equation of the tangent line, we use implicit differentiation.
Differentiate \(4(x^{2}+y^{2})^{2}=25xy^{2}\) with respect to \(x\).
Let \(u=x^{2}+y^{2}\), then the left - hand side is \(4u^{2}\). By 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 \(u = 25x\) and \(v=y^{2}\), we have \(\frac{d}{dx}(25xy^{2})=25y^{2}+50xy\frac{dy}{dx}\).
So, \(8(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=25y^{2}+50xy\frac{dy}{dx}\).
Substitute \(x = 1\) and \(y = 2\):

$$8(1^{2}+2^{2})(2\times1+2\times2\frac{dy}{dx})=25\times2^{2}+50\times1\times2\frac{dy}{dx}$$

$$8\times5(2 + 4\frac{dy}{dx})=100 + 100\frac{dy}{dx}$$

$$40(2 + 4\frac{dy}{dx})=100 + 100\frac{dy}{dx}$$

$$80+160\frac{dy}{dx}=100 + 100\frac{dy}{dx}$$

$$160\frac{dy}{dx}-100\frac{dy}{dx}=100 - 80$$

$$60\frac{dy}{dx}=20$$

$$\frac{dy}{dx}=\frac{1}{3}$$

The equation of the tangent line using the point - slope form \(y - y_{1}=m(x - x_{1})\) with \((x_{1},y_{1})=(1,2)\) and \(m=\frac{1}{3}\) is:

$$y - 2=\frac{1}{3}(x - 1)$$

$$y=\frac{1}{3}x-\frac{1}{3}+2$$

$$y=\frac{1}{3}x+\frac{5}{3}$$

Answer:

The point \((1,2)\) is on the curve \(4(x^{2}+y^{2})^{2}=25xy^{2}\). The equation of the tangent line to the curve at the point \((1,2)\) is \(y=\frac{1}{3}x+\frac{5}{3}\)