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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Straight Lines

Let the equation x(x+2)(12-k)=2 have equal roots. Then the distance...

Let the equation x(x+2)(12−k)=2\mathrm{x}(\mathrm{x}+2)(12-\mathrm{k})=2 have equal roots. Then the distance of the point (k,k2)\left(\mathrm{k}, \frac{\mathrm{k}}{2}\right) from the line 3x+4y+5=03 x+4 y+5=0 is

Options

  1. A1515
  2. B535 \sqrt{3}
  3. C15515 \sqrt{5}
  4. D1212

Correct answer

A. 1515

Step-by-step solution

(x2+2x)(12−k)=2λx2+2λx−2=0k≠12 Let 12−k=λD=04λ2+8λ=0λ=0 or  λ=−2⇒12−k=−2k=14\begin{aligned}& \left(\mathrm{x}^2+2 \mathrm{x}\right)(12-\mathrm{k})=2 \\ & \lambda \mathrm{x}^2+2 \lambda \mathrm{x}-2=0 \quad \mathrm{k} \neq 12 \text{ Let } 12-\mathrm{k}=\lambda \\ & \mathrm{D}=0 \\ & 4 \lambda^2+8 \lambda=0 \\ & \lambda=0 \text{\text{ or } } \lambda=-2 \\ & \Rightarrow 12-\mathrm{k}=-2 \\ & \mathrm{k}=14\end{aligned}\text{ So } P\left(k, \frac{k}{2}\right)=(14,7) \mathrm{d}=\left|\frac{3 \times 14+4 \times 7+5}{5}\right|=15$option (1)
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