If a curve is given by `x= a cos t + b/2 cos2t` and `y= asint + b/2 sin 2t`, then the points for which `(d^2 y)/dx^2 = 0`, are given by

Options

A. `sint=(2a^(2)+b^(2))/(3ab)`
B. `cost=-(a^(2)+2b^(2))/(3ab)`
C. `tant=a//b`
D. none of these

Answer

To solve the problem, we need to find the points for which \(\frac{d^2y}{dx^2} = 0\) for the given parametric equations:

\[ x = a \cos t + \frac{b}{2} \cos 2t \]
\[ y = a \sin t + \frac{b}{2} \sin 2t \]

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

We start by finding the first derivatives of \(x\) and \(y\):

\[ \frac{dx}{dt} = -a \sin t - b \sin 2t \]
\[ \frac{dy}{dt} = a \cos t + b \cos 2t \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule, we can express \(\frac{dy}{dx}\) in terms of \(t\):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t + b \cos 2t}{-a \sin t - b \sin 2t} \]

### Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)

To find \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(t\):

\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

First, we differentiate \(\frac{dy}{dx}\):

Using the quotient rule:

\[ \frac{d}{dt}\left(\frac{a \cos t + b \cos 2t}{-a \sin t - b \sin 2t}\right) = \frac{(-a \sin t - 2b \sin 2t)(-a \sin t - b \sin 2t) - (a \cos t + b \cos 2t)(-a \cos t - 2b \cos 2t)}{(-a \sin t - b \sin 2t)^2} \]

### Step 4: Simplify \(\frac{d^2y}{dx^2}\)

Now we will simplify the expression obtained in Step 3. After simplification, we will have:

\[ \frac{d^2y}{dx^2} = \frac{a^2 + 2b^2 + 3ab \cos t}{(a \sin t + b \sin 2t)^3} \]

### Step 5: Set \(\frac{d^2y}{dx^2} = 0\)

For \(\frac{d^2y}{dx^2} = 0\), the numerator must be zero:

\[ a^2 + 2b^2 + 3ab \cos t = 0 \]

### Step 6: Solve for \(\cos t\)

Rearranging gives:

\[ 3ab \cos t = - (a^2 + 2b^2) \]

Thus,

\[ \cos t = -\frac{a^2 + 2b^2}{3ab} \]

### Final Result

The points for which \(\frac{d^2y}{dx^2} = 0\) are given by:

\[ \cos t = -\frac{a^2 + 2b^2}{3ab} \]