The given equation is:

$4y^2 + x - 36y + 68 = 0$

Step 1: Rearrange into Conic Form

To write this equation in standard conic form, we'll complete the square for the $y$-terms.

1. Group the terms involving $y$: $4y^2 - 36y + x + 68 = 0$ Separate the $x$ and constant terms: $4(y^2 - 9y) + x + 68 = 0$

2. Complete the square for $y^2 - 9y$. First, take half of 9, square it to get $\left(\frac{9}{2}\right)^2 = \frac{81}{4}$, and add/subtract inside the parentheses. $y^2 - 9y = (y - \frac{9}{2})^2 - \frac{81}{4}$

3. Substitute this into the equation: $4 \left( (y - \frac{9}{2})^2 - \frac{81}{4} \right) + x + 68 = 0$ Simplify: $4(y - \frac{9}{2})^2 - 81 + x + 68 = 0$ $4(y - \frac{9}{2})^2 + x - 13 = 0$ Finally, isolate the constants on one side: $4(y - \frac{9}{2})^2 + x = 13$

Step 2: Identify the Conic Type

This equation is of the form: $A(y - k)^2 + x = C$ This is the equation of a parabola that opens horizontally.

Summary:

The given equation represents a parabola. In standard form, it is written as: $4(y - \frac{9}{2})^2 + x = 13$

Would you like more details or have any questions?

5 Related Questions:

1. How do you determine the direction of opening for a parabola?
2. What is the vertex of this parabola?
3. How do you convert a general conic equation to standard form?
4. What are the key features of a conic section?
5. How does completing the square help in identifying conic sections?

Tip:

When completing the square, always remember to balance the equation by adjusting the constant terms on both sides!