JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q82

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MathematicsSingle correct+4 / -1Question 82 of 90

The slope of the line touching both the parabolas $y^2=4x$ and $x^2=-32y$ is :

\dfrac18

\dfrac23

\dfrac12

\dfrac32

Answer and worked solution

Official answer: C

A tangent to $y^2=4x$ is $y=mx+\frac1m$ . It must also touch $x^2=-32y$ . Substituting, $x^2+32mx+\frac{32}m=0$ must have equal roots:
$1024m^2=\frac{128}m\implies m^3=\frac18\implies m=\frac12.$

Topics:Tangents and normals to conics Parabola

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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 82