To solve the problem using implicit differentiation and find the equation of the tangent line at the given point $\left( 0, \frac{1}{4} \right)$, follow these steps:

Step 1: Differentiate both sides implicitly

Given the equation: $x^2 + y^2 = (5x^2 + 4y^2 - x)^2$ Differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$ (so we'll apply the chain rule when differentiating terms with $y$).

Left-hand side:

$\frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx}$

Right-hand side:

Let $u(x, y) = 5x^2 + 4y^2 - x$, then: $\frac{d}{dx} \left( u^2 \right) = 2u \cdot \frac{du}{dx}$ To differentiate $u(x, y)$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx} (5x^2 + 4y^2 - x) = 10x + 8y \frac{dy}{dx} - 1$ Thus, the derivative of the right-hand side is: $2(5x^2 + 4y^2 - x) \cdot (10x + 8y \frac{dy}{dx} - 1)$

Step 2: Plug in the given point $(0, \frac{1}{4})$

Now substitute $x = 0$ and $y = \frac{1}{4}$ into the differentiated equation to solve for $\frac{dy}{dx}$.

First, evaluate the function $u(x, y)$ at $(0, \frac{1}{4})$: $u(0, \frac{1}{4}) = 5(0)^2 + 4\left(\frac{1}{4}\right)^2 - 0 = \frac{4}{16} = \frac{1}{4}$

Now, substitute this into the full derivative equation: $2(0) + 2\left(\frac{1}{4}\right)\frac{dy}{dx} = 2\left(\frac{1}{4}\right) \cdot (10(0) + 8\left(\frac{1}{4}\right)\frac{dy}{dx} - 1)$

Simplifying this: $\frac{1}{2} \frac{dy}{dx} = \frac{1}{2} \cdot (2\frac{dy}{dx} - 1)$

Multiply both sides by 2: $\frac{dy}{dx} = 2\frac{dy}{dx} - 1$

Rearranging: $\frac{dy}{dx} - 2\frac{dy}{dx} = -1$ $-\frac{dy}{dx} = -1$ $\frac{dy}{dx} = 1$

Step 3: Write the equation of the tangent line

The slope of the tangent line at the point $\left( 0, \frac{1}{4} \right)$ is 1. Therefore, the equation of the tangent line in point-slope form is: $y - \frac{1}{4} = 1(x - 0)$ Simplifying: $y = x + \frac{1}{4}$

Final Answer:

The equation of the tangent line to the curve at the point $\left( 0, \frac{1}{4} \right)$ is: $y = x + \frac{1}{4}$

Would you like further details or have any questions?

Related Questions:

1. How is implicit differentiation different from explicit differentiation?
2. Can you use the same process for other curves defined by implicit equations?
3. What would change if the point were different, say $(1, 0)$?
4. How would you interpret the tangent line geometrically for a cardioid?
5. What is the physical meaning of the slope of the tangent line at a point on a curve?

Tip:

When performing implicit differentiation, always keep track of when to apply the chain rule, especially for terms involving $y$ as a function of $x$.