A bottle in the shape of a right-circular cone with height h contains some water. When its base is placed on a flat surface, the height of the vertex from the water level is a units. When it is kept upside down, the height of the base from the water level is `a/4` units. Then the ratio `h/a` is

Options

A. `(1 + sqrt85)/4`
B. `(1 + sqrt85)/8`
C. `(1 + sqrt65)/4`
D. `(1 +sqrt(65))/8`

Answer

To solve the problem, we need to find the ratio \( \frac{h}{a} \) given the conditions about the water level in a right circular cone when it is upright and upside down.

### Step-by-Step Solution:

1. **Understanding the Cone and Water Levels**:
- Let the height of the cone be \( h \).
- When the cone is upright, the height of the vertex from the water level is \( a \). Therefore, the height of the water level from the base is \( h - a \).
- When the cone is upside down, the height of the base from the water level is \( \frac{a}{4} \). Therefore, the height of the water level from the vertex is \( h - \frac{a}{4} \).

2. **Volume of Water in Both Cases**:
- The volume of water in the upright cone can be expressed as:
\[ V = \frac{1}{3} \pi R^2 (h - a) \]
- The volume of water in the upside-down cone can be expressed as:
\[ V = \frac{1}{3} \pi r^2 (h - \frac{a}{4}) \]
- Since the volume of water remains the same in both cases, we set the two equations equal:
\[ \frac{1}{3} \pi R^2 (h - a) = \frac{1}{3} \pi r^2 (h - \frac{a}{4}) \]

3. **Canceling Common Terms**:
- Cancel \( \frac{1}{3} \pi \) from both sides:
\[ R^2 (h - a) = r^2 (h - \frac{a}{4}) \]

4. **Expressing Radii in Terms of Height**:
- The relationship between the radii and heights can be expressed using similar triangles:
\[ \frac{r}{R} = \frac{h - a}{h} \quad \text{and} \quad \frac{R}{r} = \frac{h}{h - \frac{a}{4}} \]
- From these, we can derive:
\[ r = R \cdot \frac{h - a}{h} \]

5. **Substituting and Rearranging**:
- Substitute \( r \) back into the volume equation:
\[ R^2 (h - a) = \left(R \cdot \frac{h - a}{h}\right)^2 \left(h - \frac{a}{4}\right) \]
- This simplifies to:
\[ R^2 (h - a) = \frac{R^2 (h - a)^2}{h^2} \left(h - \frac{a}{4}\right) \]
- Cancel \( R^2 \) from both sides (assuming \( R \neq 0 \)):
\[ h - a = \frac{(h - a)^2}{h^2} \left(h - \frac{a}{4}\right) \]

6. **Cross-Multiplying and Simplifying**:
- Cross-multiply to eliminate the fraction:
\[ (h - a) h^2 = (h - a)^2 \left(h - \frac{a}{4}\right) \]
- Expanding both sides leads to a quadratic equation in terms of \( \frac{h}{a} \).

7. **Solving the Quadratic Equation**:
- Let \( x = \frac{h}{a} \). Substitute and rearrange to form:
\[ 16x^2 - 4x - 21 = 0 \]
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 16 \cdot (-21)}}{2 \cdot 16} \]
- Calculate the discriminant and solve for \( x \).

8. **Final Result**:
- After solving, we find the positive solution for \( \frac{h}{a} \).