JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
The equation of a common tangent to the parabolas y=x² 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)
Answer and solution
Answer (b) y=4(x-1)
Solution
Maths sir · 4 steps
Write the line as y = mx + c and make it touch each parabola: two zero discriminants give m and c.
- Touch the first. Put y = mx + c into y = x² to get x² - mx - c = 0. The line touches the parabola when this quadratic has one repeated root, so m² + 4c = 0, that is c = -m²/4.
- Touch the second. Put y = mx + c into y = -(x-2)² to get x² + (m-4)x + (c+4) = 0. Its discriminant must also be zero: (m-4)² - 4(c+4) = 0.
(m-4)² - 4(4 - m²/4) = 0 ⇒ 2m² - 8m = 0 ⇒ m = 0 or m = 4
- Pick the line. m = 0 gives c = 0, which is the line y = 0. m = 4 gives c = -4, which is the line y = 4x - 4. This second line touches y = x² at (2, 4) and touches y = -(x-2)² at (0, -4).
- So. The common tangent among the options is y = 4x - 4 = 4(x-1), which is option (b).
The idea to keep. For a common tangent, take y = mx + c, substitute into each curve and set each discriminant to zero. Solve the two conditions together for m and c.
Answer from the NTA final answer key. Solution by PrepWiser.