Question

The equation of a common tangent to the parabolas \( y = x^2 \) and \( y = -(x - 2)^2 \) is:

• A. \( y = 4(x - 2) \)
• B. \( y = 4(x - 1) \)
• C. \( y = 4(x + 1) \)
• D. \( y = 4(x + 2) \)

Hint:

For finding the common tangent to two parabolas, use the equation of the tangent in slope form and ensure that the discriminant is zero for the quadratic equation formed.

Answer 1

The Correct Option is B

The equation of the tangent to the parabola \( y = x^2 \) at any point \( (t, t^2) \) is given by: \[ tx = y + at^2 \] Since \( y = x^2 \), we substitute and rewrite: \[ y = tx - \frac{t^2}{4} \] Solving this with the equation \( y = (x - 2)^2 \), we equate: \[ tx - \frac{t^2}{4} = (x - 2)^2 \] Expanding and simplifying: \[ x^2 + x(t - 4) - \frac{t^2}{4} + 4 = 0 \] For a common tangent, the discriminant must be zero: \[ (t - 4)^2 - 4 \left( 4 - \frac{t^2}{4} \right) = 0 \] Solving for \( t \): \[ t^2 - 4t = 0 \Rightarrow t = 0 \text{ or } t = 4 \] Substituting \( t = 4 \) in the tangent equation: \[ y = 4(x - 1) \] Thus, the correct answer is \(\boxed{(b)}\).