The sum of the squares of the perpendicular drawn from the points (0,1) and `(0,-1)` to any tangent to a curve is 2. The equation of the curve, is

Options

A. `2y=C(x+2)`
B. `y=C(x+1)`
C. `y=C(x+2)`
D. `y=C(x+2)`

Explanation

To solve the problem, we need to find the equation of the curve such that the sum of the squares of the perpendiculars drawn from the points (0, 1) and (0, -1) to any tangent to the curve is equal to 2.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given two points (0, 1) and (0, -1) and need to find a curve such that the sum of the squares of the distances from these points to any tangent line of the curve equals 2.

2. **Equation of the Tangent Line**:
Let the equation of the tangent line to the curve at a point (x1, y1) be given by:
\[ y - y_1 = \frac{dy}{dx}(x - x_1) \]
Rearranging gives:
\[ y = \frac{dy}{dx}x - \frac{dy}{dx}x_1 + y_1 \]

3. **Finding the Distance from Points to the Tangent**:
The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our tangent line, we can rewrite it in the form:
\[ -\frac{dy}{dx}x + y - (y_1 - \frac{dy}{dx}x_1) = 0 \]
Here, \(A = -\frac{dy}{dx}\), \(B = 1\), and \(C = -(y_1 - \frac{dy}{dx}x_1)\).

4. **Calculating Distances**:
For the point (0, 1):
\[ d_1 = \frac{|- \frac{dy}{dx}(0) + 1 - (y_1 - \frac{dy}{dx}x_1)|}{\sqrt{(-\frac{dy}{dx})^2 + 1^2}} = \frac{|1 - y_1 + \frac{dy}{dx}x_1|}{\sqrt{(\frac{dy}{dx})^2 + 1}} \]
For the point (0, -1):
\[ d_2 = \frac{|- \frac{dy}{dx}(0) - 1 - (y_1 - \frac{dy}{dx}x_1)|}{\sqrt{(-\frac{dy}{dx})^2 + 1^2}} = \frac{|-1 - y_1 + \frac{dy}{dx}x_1|}{\sqrt{(\frac{dy}{dx})^2 + 1}} \]

5. **Setting Up the Equation**:
According to the problem, the sum of the squares of these distances equals 2:
\[ d_1^2 + d_2^2 = 2 \]
Substituting \(d_1\) and \(d_2\) into this equation leads to a complex expression.

6. **Simplifying the Expression**:
After some algebraic manipulation, we can derive that:
\[ (x \frac{dy}{dx} - y + 1)^2 + (x \frac{dy}{dx} - y - 1)^2 = 2 \]
This can be simplified to yield a relationship involving \(y\) and \(x\).

7. **Final Integration**:
After simplifying, we find that:
\[ \frac{dy}{dx} = \frac{y}{x \pm 1} \]
Integrating both sides gives:
\[ \log |y| = \log |x \pm 1| + \log |C| \]
Thus, we arrive at:
\[ y = C(x \pm 1) \]

8. **Conclusion**:
The required equation of the curve is:
\[ y = C(x + 1) \quad \text{or} \quad y = C(x - 1) \]
where \(C\) is a constant.