JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

If $y=m_1x+c_1$ and $y=m_2x+c_2$ , $m_1\ne m_2$ are two common tangents of circle $x^2+y^2=2$ and parabola $y^2=x$ , then the value of $8|m_1m_2|$ is equal to :

3+4\sqrt2

-5+6\sqrt2

-4+3\sqrt2

7+6\sqrt2

Answer and worked solution

Official answer: C

Tangent to $y^2=x$ : $y=mx+\dfrac1{4m}$ . Tangent to the circle: $c^2=2(1+m^2)$ .

Equation. $\dfrac1{16m^2}=2+2m^2$ gives $32m^4+32m^2-1=0$ , so $m^2=\dfrac{-4+3\sqrt2}8$ .

Slopes are $\pm m$ , so $8|m_1m_2|=8m^2=-4+3\sqrt2$ .

(C)

Topics:Tangents and normals to conics Tangents and normals

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 16