The value of k such that the family of parabolas `y=cx^(2)+k` is the orthogonal trajectory of the family of ellipse `x^(2)+2y^(2)-y=c,` is

Options

A. `1`
B. `1/2`
C. `1/3`
D. `1/4`

Answer

To find the value of \( k \) such that the family of parabolas \( y = cx^2 + k \) is the orthogonal trajectory of the family of ellipses given by \( x^2 + 2y^2 - y = c \), we will follow these steps:

### Step 1: Differentiate the equation of the ellipse
We start with the equation of the ellipse:
\[ x^2 + 2y^2 - y = c \]
Differentiating both sides with respect to \( x \) gives:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) - \frac{d}{dx}(y) = \frac{d}{dx}(c) \]
This simplifies to:
\[ 2x + 4y \frac{dy}{dx} - \frac{dy}{dx} = 0 \]
Rearranging terms, we have:
\[ 2x + (4y - 1) \frac{dy}{dx} = 0 \]
Thus, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{2x}{4y - 1} \]

### Step 2: Differentiate the equation of the parabola
Next, we differentiate the equation of the parabola:
\[ y = cx^2 + k \]
Differentiating gives:
\[ \frac{dy}{dx} = 2cx \]

### Step 3: Set up the orthogonality condition
For the curves to be orthogonal, the product of their slopes must equal \(-1\):
\[ \left(-\frac{2x}{4y - 1}\right)(2cx) = -1 \]
This simplifies to:
\[ -\frac{4cx^2}{4y - 1} = -1 \]
Thus, we have:
\[ \frac{4cx^2}{4y - 1} = 1 \]
Cross-multiplying gives:
\[ 4cx^2 = 4y - 1 \]

### Step 4: Substitute the equation of the parabola into the orthogonality condition
Now, substitute \( y = cx^2 + k \) into the equation:
\[ 4cx^2 = 4(cx^2 + k) - 1 \]
Expanding the right side:
\[ 4cx^2 = 4cx^2 + 4k - 1 \]
Subtracting \( 4cx^2 \) from both sides results in:
\[ 0 = 4k - 1 \]

### Step 5: Solve for \( k \)
Rearranging gives:
\[ 4k = 1 \quad \Rightarrow \quad k = \frac{1}{4} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{4}} \]