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

The given equation is \(x(x+2)(12-k) = 2\). It is mentioned that the equation has equal roots. For an equation to have equal roots, its discriminant should be zero. The given equation can be rearranged as a quadratic in \(x\):

\(x^2(12-k) + 2x(12-k) - 2 = 0\).

The coefficients of this equation are:

• \(a = 12-k\)
• \(b = 2(12-k)\)
• \(c = -2\)

For equal roots, the discriminant \(\Delta\) should be zero:

\(\Delta = b^2 - 4ac = 0\)

Substitute the values of \(a\), \(b\), and \(c\):

\((2(12-k))^2 - 4(12-k)(-2) = 0\)

\(4(12-k)^2 + 8(12-k) = 0\)

\(4(12-k)((12-k)+2) = 0\)

This simplifies to:

\(4(12-k)(14-k) = 0\)

Thus, \(12-k = 0\) or \(14-k = 0\). Solving these, we find:

• \(k = 12\)
• \(k = 14\)

Next, we find the distance of the point \(\left(k, \frac{k}{2}\right)\) from the line \(3x + 4y + 5 = 0\).

The distance \(D\) from a point \((x_1, y_1)\) to a line \(Ax+By+C=0\) is given by:

\(D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

Let's calculate this for the different values of \(k\):

1. When \(k = 12\), the point is \((12, 6)\):

\(D = \frac{|3 \cdot 12 + 4 \cdot 6 + 5|}{\sqrt{3^2 + 4^2}}\)

\(D = \frac{|36 + 24 + 5|}{\sqrt{9 + 16}}\)

\(D = \frac{65}{5} = 13\)

1. When \(k = 14\), the point is \((14, 7)\):

\(D = \frac{|3 \cdot 14 + 4 \cdot 7 + 5|}{\sqrt{3^2 + 4^2}}\)

\(D = \frac{|42 + 28 + 5|}{\sqrt{25}}\)

\(D = \frac{75}{5} = 15\)

Thus, the correct option considering both scenarios is 15.

Answer 2

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

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

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

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

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