JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$, then $12m^{2}$ is equal to :
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Solution
Formulas used. (1) $y=mx+c$ touches $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ iff $c^2=a^2m^2+b^2$(2) $y=mx+c$ touches $x^2+y^2=r^2$ iff $c^2=r^2(1+m^2)$Step 1 - Use (1). $c^2=16m^2+9$. Step 2 - Use (2). $c^2=12+12m^2$. Step 3 - Equate. $16m^2+9=12+12m^2\Rightarrow4m^2=3\Rightarrow m^2=\frac34$. Step 4 - Compute. $12m^2=9$. Answer: B