Question

The equation of tangent to the curves \(x = 1-3t^2\) and \(y = t-3t^3\) at the point \((-2,2)\) is...

• A. \(4x+3y+2 = 0\)
• B. \(4x-3y+2 = 0\)
• C. \(3x+4y+2 = 0\)
• D. \(3x-4y+2 = 0\)

Hint:

Find the parameter value at the point, then dy/dx = (dy/dt)/(dx/dt).

Answer 1

The Correct Option is A

Step 1: Check the point on options
Substitute $(-2,2)$: option (A): $-8+6+2 = 0$ holds. Option (B): $-8-6+2 \ne 0$. Option (C): $-6+8+2\ne0$. Option (D): $-6-8+2\ne0$.

Step 2: Check slope
With parameter $t=-1$, slope $= -\frac43$, and option (A) has slope $-\frac43$. Option (A).

Final Answer:
4x + 3y + 2 = 0. \[ \boxed{\text{(A)}\ 4x+3y+2=0} \]