Tangents are drawn from the origin to the curve `y=cos X`. Their points of contact lie on

Options

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

Answer

To find the points of contact of the tangents drawn from the origin to the curve \( y = \cos x \), we can follow these steps:

### Step 1: Define the point of contact
Let the point of contact on the curve be \( (x_1, y_1) \). Since the point lies on the curve \( y = \cos x \), we have:
\[ y_1 = \cos x_1 \]

### Step 2: Find the derivative of the curve
The derivative of the curve \( y = \cos x \) is:
\[ \frac{dy}{dx} = -\sin x \]
At the point \( (x_1, y_1) \), the slope of the tangent line is:
\[ \frac{dy}{dx} \bigg|_{(x_1, y_1)} = -\sin x_1 \]

### Step 3: Write the equation of the tangent line
The equation of the tangent line at the point \( (x_1, y_1) \) can be expressed as:
\[ y - y_1 = \frac{dy}{dx} (x - x_1) \]
Substituting the values, we get:
\[ y - \cos x_1 = -\sin x_1 (x - x_1) \]

### Step 4: Substitute the origin into the tangent equation
Since the tangent line passes through the origin \( (0, 0) \), we substitute \( x = 0 \) and \( y = 0 \):
\[ 0 - \cos x_1 = -\sin x_1 (0 - x_1) \]
This simplifies to:
\[ -\cos x_1 = x_1 \sin x_1 \]
Rearranging gives us:
\[ \cos x_1 = -x_1 \sin x_1 \]

### Step 5: Use the identity \( \sin^2 x + \cos^2 x = 1 \)
We know that \( y_1 = \cos x_1 \), so we can square both sides:
\[ \cos^2 x_1 = y_1^2 \]
Substituting this into our equation gives:
\[ y_1^2 = -x_1 \sin x_1 \]

### Step 6: Relate \( \sin^2 x_1 \) and \( \cos^2 x_1 \)
Using the identity \( \sin^2 x_1 + \cos^2 x_1 = 1 \):
\[ \sin^2 x_1 = 1 - \cos^2 x_1 = 1 - y_1^2 \]
Substituting this into the equation:
\[ y_1^2 + \frac{x_1^2}{y_1^2} = 1 \]

### Step 7: Final equation
After simplification, we arrive at:
\[ \frac{1}{y_1^2} = 1 + \frac{1}{x_1^2} \]
Replacing \( x_1 \) and \( y_1 \) with \( x \) and \( y \) respectively, we get:
\[ \frac{1}{y^2} = 1 + \frac{1}{x^2} \]

### Conclusion
Thus, the points of contact of the tangents from the origin to the curve \( y = \cos x \) lie on the curve defined by:
\[ \frac{1}{y^2} = 1 + \frac{1}{x^2} \]