JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q11

MathematicsSingle correct+4 / -1Question 11 of 90

If $m$ is the slope of a common tangent to the curves $\dfrac{x^2}{16}+\dfrac{y^2}9=1$ and $x^2+y^2=12$ , then $12m^2$ is equal to :

Answer and worked solution

Official answer: B

Tangency conditions for $y=mx+c$ : ellipse $c^2=16m^2+9$ , circle $c^2=12(1+m^2)$ .

Equate. $4m^2=3$ , so $12m^2=9$ .

(B)

Topics:Tangents and normals to conics Tangents and normals

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 11