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Common Tangent to y = x^2 and y = x^2 - 2x + 2 | JEE Advanced Parabola

JEE Maths question with a full step-by-step solution.

Question

The two parabolas

y = x^{2}

and

y = x^{2}-2x+2

have a common tangent. Find its equation.

y = 2x-1

4y - x + 4 = 0

4x - 4y - 1 = 0

correct

y + 4x + 4 = 0

Solution

Step 1: No vertical line touches

y = x^{2}

, so a common tangent has the form

y = mx + c

. Putting it in

y = x^{2}

x^{2} - mx - c = 0

Tangency means equal roots, so the discriminant vanishes:

m^{2} + 4c = 0 \ \Longrightarrow\ c = -\frac{m^{2}}{4}

Hence every tangent to

y=x^{2}

y = mx - \frac{m^{2}}{4}

Step 2: The same line touches

y = x^{2}-2x+2

mx - \frac{m^{2}}{4} = x^{2}-2x+2 \ \Longrightarrow\ x^{2} - (2+m)x + \left(2 + \frac{m^{2}}{4}\right) = 0

Step 3: This discriminant is zero as well:

(m+2)^{2} - 4\left(2 + \frac{m^{2}}{4}\right) = 0

m^{2} + 4m + 4 - 8 - m^{2} = 0 \ \Longrightarrow\ 4m - 4 = 0 \ \Longrightarrow\ m = 1

m

is unique, so there is exactly one common tangent. Step 4:

c = -\frac{1}{4}, \qquad y = x - \frac14 \ \Longrightarrow\ 4x - 4y - 1 = 0

Answer: (3).

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