Tangents are drawn from the origin to the curve `y = sin x`. Prove that their points of contact lie on the curve `x^(2) y^(2) = (x^(2) - y^(2))`

Options

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

Answer

To prove that the points of contact of tangents drawn from the origin to the curve \( y = \sin x \) lie on the curve \( x^2 y^2 = x^2 - y^2 \), we can follow these steps:

### Step 1: Equation of the Tangent
The equation of a line (tangent) passing through the origin can be expressed as:
\[ y = mx \]
where \( m \) is the slope of the tangent.

### Step 2: Point of Contact
Let the point of contact of the tangent with the curve \( y = \sin x \) be \( P(x_1, y_1) \). Since this point lies on the curve, we have:
\[ y_1 = \sin x_1 \]

### Step 3: Slope of the Tangent
To find the slope \( m \) at the point of contact, we need to differentiate \( y = \sin x \):
\[ \frac{dy}{dx} = \cos x \]
At the point \( P(x_1, y_1) \), the slope of the tangent is:
\[ m = \cos x_1 \]

### Step 4: Relating Slope to Coordinates
Since the point \( P(x_1, y_1) \) lies on the line \( y = mx \), we can substitute \( y_1 \) and \( m \):
\[ y_1 = m x_1 \implies \sin x_1 = \cos x_1 \cdot x_1 \]

### Step 5: Squaring Both Sides
Now, squaring both sides gives:
\[ \sin^2 x_1 = (x_1 \cos x_1)^2 \]
This can be expanded as:
\[ \sin^2 x_1 = x_1^2 \cos^2 x_1 \]

### Step 6: Using the Pythagorean Identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can substitute \( \cos^2 x_1 \):
\[ \sin^2 x_1 = x_1^2 (1 - \sin^2 x_1) \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ \sin^2 x_1 + x_1^2 \sin^2 x_1 = x_1^2 \]
Factoring out \( \sin^2 x_1 \):
\[ \sin^2 x_1 (1 + x_1^2) = x_1^2 \]

### Step 8: Final Form
Now, we can express this as:
\[ x_1^2 \sin^2 x_1 = x_1^2 - \sin^2 x_1 \]
This is equivalent to:
\[ x_1^2 y_1^2 = x_1^2 - y_1^2 \]
where \( y_1 = \sin x_1 \).

### Conclusion
Thus, we have shown that the points of contact of the tangents drawn from the origin to the curve \( y = \sin x \) lie on the curve \( x^2 y^2 = x^2 - y^2 \).

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