ABCD and EFGC are squares and the curve `y=ksqrtx` passes through the origin D and the points B and F.The ratio of `(FG)/(BC)` is:

Options

A. A. `(sqrt3+1)/4`
B. B. `(sqrt3+1)/2`
C. C. `(sqrt5+1)/4`
D. D. `(sqrt5+1)/2`

Answer

To solve the problem, we need to find the ratio of the lengths \( FG \) and \( BC \) where \( ABCD \) and \( EFGC \) are squares and the curve \( y = k\sqrt{x} \) passes through points \( D \), \( B \), and \( F \).

### Step-by-step Solution:

1. **Understanding the Curve**:
The curve is given by the equation \( y = k\sqrt{x} \). This can be rewritten in a standard parabolic form by squaring both sides:
\[ y^2 = k^2 x \]
This indicates that the parabola opens to the right.

2. **Identifying Points**:
Since \( D \) is at the origin, we have \( D(0, 0) \). Let’s assume the coordinates of point \( B \) on the parabola are \( (t_1^2, 2a t_1) \) and point \( F \) has coordinates \( (t_2^2, 2a t_2) \).

3. **Finding \( a \)**:
From the parabola, we know that \( a = \frac{k^2}{4} \). Thus, the coordinates of \( B \) become:
\[ B\left(t_1^2, \frac{k^2}{2} t_1\right) \]
and for \( F \):
\[ F\left(t_2^2, \frac{k^2}{2} t_2\right) \]

4. **Finding Lengths of the Square**:
Since \( ABCD \) is a square, the length of \( BC \) (which is equal to \( CD \)) can be calculated from the coordinates of \( B \):
\[ BC = \frac{k^2}{2} t_1 \]
Similarly, for square \( EFGC \), the length \( FG \) can be expressed as:
\[ FG = \frac{k^2}{2} t_2 \]

5. **Setting Up the Ratio**:
We need to find the ratio \( \frac{FG}{BC} \):
\[ \frac{FG}{BC} = \frac{\frac{k^2}{2} t_2}{\frac{k^2}{2} t_1} = \frac{t_2}{t_1} \]

6. **Finding \( t_1 \) and \( t_2 \)**:
From the properties of the squares and the coordinates, we can derive that:
\[ t_1 = 2 \quad \text{(from the calculations)} \]
For \( t_2 \), we solve the quadratic equation derived from the geometry of the squares:
\[ t_2^2 - 2t_2 - 4 = 0 \]
Using the quadratic formula:
\[ t_2 = \frac{2 \pm \sqrt{4 + 16}}{2} = 1 \pm \sqrt{5} \]
Since we are in the first quadrant, we take \( t_2 = 1 + \sqrt{5} \).

7. **Calculating the Ratio**:
Now substituting back into the ratio:
\[ \frac{FG}{BC} = \frac{1 + \sqrt{5}}{2} \]

### Final Answer:
Thus, the ratio \( \frac{FG}{BC} \) is:
\[ \frac{FG}{BC} = \frac{1 + \sqrt{5}}{2} \]