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The problem asks to find the equation of the tangent to the curve:

\[
y^2 + 2\sqrt{xy} = 21 - x^2
\]

at the point \( (1, 4) \).

### Solution:

1. **Implicit Differentiation**:
   The given equation is \( y^2 + 2\sqrt{xy} = 21 - x^2 \). To find the slope of the tangent line at the point \( (1, 4) \), we need to differentiate both sides of the equation with respect to \( x \), using implicit differentiation.

   - Differentiating \( y^2 \) with respect to \( x \):  
     \[
     \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}
     \]
     
   - Differentiating \( 2\sqrt{xy} \) using the product rule and chain rule:  
     \[
     \frac{d}{dx}(2\sqrt{xy}) = 2 \cdot \frac{1}{2\sqrt{xy}} \left( y + x \frac{dy}{dx} \right) = \frac{y + x\frac{dy}{dx}}{\sqrt{xy}}
     \]
     
   - Differentiating \( 21 - x^2 \) with respect to \( x \):  
     \[
     \frac{d}{dx}(21 - x^2) = -2x
     \]

2. **Combine all terms**:
   After differentiating, we get the equation:
   \[
   2y \frac{dy}{dx} + \frac{y + x \frac{dy}{dx}}{\sqrt{xy}} = -2x
   \]

3. **Substitute the point (1, 4)**:
   At \( (x, y) = (1, 4) \), substitute \( x = 1 \) and \( y = 4 \) into the equation:
   \[
   2(4) \frac{dy}{dx} + \frac{4 + 1 \cdot \frac{dy}{dx}}{\sqrt{1 \cdot 4}} = -2(1)
   \]
   Simplifying:
   \[
   8 \frac{dy}{dx} + \frac{4 + \frac{dy}{dx}}{2} = -2
   \]

4. **Solve for \( \frac{dy}{dx} \)**:
   Multiply through by 2 to eliminate the denominator:
   \[
   16 \frac{dy}{dx} + 4 + \frac{dy}{dx} = -4
   \]
   Combine like terms:
   \[
   17 \frac{dy}{dx} = -8
   \]
   Solving for \( \frac{dy}{dx} \):
   \[
   \frac{dy}{dx} = \frac{-8}{17}
   \]

5. **Find the equation of the tangent line**:
   The slope of the tangent line at the point \( (1, 4) \) is \( \frac{-8}{17} \). Using the point-slope form of the equation of a line:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m = \frac{-8}{17} \) and \( (x_1, y_1) = (1, 4) \), we get:
   \[
   y - 4 = \frac{-8}{17}(x - 1)
   \]
   Simplifying this equation:
   \[
   y = \frac{-8}{17}x + \frac{8}{17} + 4
   \]
   \[
   y = \frac{-8}{17}x + \frac{76}{17}
   \]

Thus, the equation of the tangent line is:
\[
y = \frac{-8}{17}x + \frac{76}{17}
\]

Would you like more details on any step? Here are five related questions you could explore:

1. What is implicit differentiation, and how is it applied?
2. How does the product rule for derivatives work?
3. How would you find the tangent if the curve were parametric?
4. How does one derive the general form of the equation of a tangent line?
5. What are other ways to approximate the slope of a curve at a point?

**Tip**: Always check if the point lies on the curve before computing the tangent line.

Find the equation of the tangent to the curve y^2 + 2√xy = 21 − x^2 at the point (1, 4).

Learn how to find the equation of the tangent line to the curve y^2 + 2√xy = 21 − x^2 at the point (1, 4) using implicit differentiation. This problem covers key calculus concepts such as implicit differentiation, the product rule, and the chain rule. Suitable for high school calculus students.

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