Question

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

• A. 15
• B. \( 5\sqrt{5} \)
• C. \( 15\sqrt{5} \)
• D. 12

Hint:

For a quadratic equation to have equal roots, its discriminant must be zero. Use this condition to find the value of \( k \). Once \( k \) is found, substitute the point \( \left(k, \frac{k}{2}\right) \) into the formula for the distance of a point from a line.

Answer 1

The Correct Option is A

\[\begin{align*} (x^2 + 2x)(12 - k) &= 2 \\ \text{Let } \lambda &= 12 - k \quad \Rightarrow \quad (x^2 + 2x)\lambda = 2 \\ \Rightarrow \lambda x^2 + 2\lambda x - 2 &= 0 \qquad \text{(Quadratic in } x \text{, valid if } k \ne 12 \text{)} \\ \text{Discriminant: } D &= (2\lambda)^2 + 4\lambda \cdot 2 = 4\lambda^2 + 8\lambda \\ \text{Set } D = 0 \text{ for equal roots:} \\ 4\lambda^2 + 8\lambda &= 0 \\ \Rightarrow \lambda( \lambda + 2 ) &= 0 \\ \Rightarrow \lambda = 0 \text{ or } \lambda = -2 \end{align*}\]

\[\begin{align*} \text{If } \lambda = -2, \text{ then } 12 - k &= -2 \Rightarrow k = 14 \end{align*}\]

\[ \therefore P(k) = \left(14, \frac{7}{2} \right) \]

\( \text{Now calculate } d = \frac{3 \times 14 + 4 \times 7 + 5}{5} = \frac{42 + 28 + 5}{5} = \frac{75}{5} = 15 \)

\( \text{Correct option: (1)} \)