Question

The equation of the tangent at any point of the curve \[ x = a \cos 2t, \, y = 2 \sqrt{2a} \sin t \, \text{with} \, m \, \text{as its slope is} \]

• A. \( y = mx + a \left( \frac{1}{m} - \frac{1}{m} \right) \)
• B. \( y = mx - a \left( \frac{m + 1}{m} \right) \)
• C. \( y = mx + a \left( \frac{a + 1}{a} \right) \)
• D. \( y = amx + a \left( \frac{m - 1}{m} \right) \)

Hint:

To find the equation of a tangent for a parametric curve, first calculate \( \frac{dy}{dx} \), then use the general formula for the tangent: \( y - y_1 = m(x - x_1) \).

Answer 1

The Correct Option is B

Given the parametric equations of the curve: \[ x = a \cos 2t, \, y = 2 \sqrt{2a} \sin t \] The slope of the tangent is denoted by \( m \). We need to find the equation of the tangent at any point on this curve. To do so, we first compute the derivative \( \frac{dy}{dx} \) (which represents the slope of the tangent). 

Step 1: Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) - The derivative of \( x = a \cos 2t \) with respect to \( t \) is: \[ \frac{dx}{dt} = -2a \sin 2t \] - The derivative of \( y = 2 \sqrt{2a} \sin t \) with respect to \( t \) is: \[ \frac{dy}{dt} = 2 \sqrt{2a} \cos t \] 

Step 2: Find \( \frac{dy}{dx} \) To find the slope \( m \) of the tangent, we use the formula: \[ m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] Substituting the values: \[ m = \frac{2 \sqrt{2a} \cos t}{-2a \sin 2t} \] Since \( \sin 2t = 2 \sin t \cos t \), we can simplify the expression as: \[ m = \frac{2 \sqrt{2a} \cos t}{-4a \sin t \cos t} = -\frac{\sqrt{2a}}{2a \sin t} \] 

Step 3: Find the equation of the tangent The general form of the equation of the tangent to the curve is: \[ y - y_1 = m(x - x_1) \] Here, \( (x_1, y_1) \) is the point on the curve at any given value of \( t \), and \( m \) is the slope of the tangent at that point. Substitute \( x_1 = a \cos 2t \) and \( y_1 = 2 \sqrt{2a} \sin t \) into the equation. After some simplification, we get the equation of the tangent as: \[ y = mx - a \left( \frac{m + 1}{m} \right) \] Thus, the equation of the tangent is \( \boxed{y = mx - a \left( \frac{m + 1}{m} \right)} \). This corresponds to option \( \boxed{(b)} \).