JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q12

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 12 of 90

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

y=4(x-2)

y=4(x-1)

y=4(x+1)

y=4(x+2)

Answer and worked solution

Official answer: B

Tangent to $y=x^2$ : $y=mx-\frac{m^2}4$ .

Touching the second parabola. Setting $mx-\frac{m^2}4=-(x-2)^2$ gives $x^2+(m-4)x+4-\frac{m^2}4=0$ , with discriminant $2m^2-8m=0$ : $m=0$ or $m=4$ .

$m=4$ gives $y=4(x-1)$ . (B)

Topics:Parabola Tangents and normals to conics

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 12