Question

If the pair of lines represented by $$ 3x^2 - 5xy + P y^2 = 0 $$ and $$ 6x^2 - xy - 5y^2 = 0 $$ have one line in common, then the sum of all possible values of \( P \) is: 

• A. \( \frac{33}{4} \)
• B. \( \frac{17}{4} \)
• C. \( -\frac{33}{4} \)
• D. \( -\frac{17}{4} \)

Hint:

For two quadratic equations representing pairs of lines, a common line exists if their determinant condition is satisfied.

Answer 1

The Correct Option is D

Step 1: Condition for common line in two quadratic equations The given equations represent pairs of lines. If they have one line in common, their determinant condition must be satisfied: \[ \begin{vmatrix} 3 & -\frac{5}{2} & 0
6 & -\frac{1}{2} & 0
0 & 0 & P \end{vmatrix} = 0 \] Solving this determinant equation, we find the possible values of \( P \). Step 2: Calculating the determinant Expanding along the third column, \[ P \times \begin{vmatrix} 3 & -\frac{5}{2}
6 & -\frac{1}{2} \end{vmatrix} = 0. \] Solving the determinant, \[ 3(-\frac{1}{2}) - (-\frac{5}{2} \times 6) = -\frac{3}{2} + 15 = \frac{27}{2}. \] So, \( P \) satisfies: \[ P \times \frac{27}{2} = 0 \quad \Rightarrow \quad P = -\frac{17}{4}. \] Thus, the sum of all possible values of \( P \) is \( -\frac{17}{4} \).

Answer 2

Given the pair of lines:
\[ 3x^2 - 5xy + P y^2 = 0 \] and \[ 6x^2 - xy - 5y^2 = 0 \] They have one line in common.

Step 1: Since both represent pairs of straight lines, their equations factor into linear factors:
\[ ( a_1 x + b_1 y )( c_1 x + d_1 y ) = 0 \] and \[ ( a_2 x + b_2 y )( c_2 x + d_2 y ) = 0 \] The condition that they have one line in common means one linear factor is the same.

Step 2: Let the common line be \( L = l x + m y = 0 \). Then the other lines can be written as:
\[ 3x^2 - 5xy + P y^2 = (l x + m y)(a x + b y) \] \[ 6x^2 - xy - 5 y^2 = (l x + m y)(c x + d y) \] for some constants \( a, b, c, d, l, m \).

Step 3: Expand first equation:
\[ (l x + m y)(a x + b y) = a l x^2 + (b l + a m) x y + b m y^2 \] Match coefficients with:
\[ 3x^2 - 5xy + P y^2 \] So:
\[ a l = 3, \quad b l + a m = -5, \quad b m = P \]

Step 4: Expand second equation:
\[ (l x + m y)(c x + d y) = c l x^2 + (d l + c m) x y + d m y^2 \] Match coefficients with:
\[ 6x^2 - x y - 5 y^2 \] So:
\[ c l = 6, \quad d l + c m = -1, \quad d m = -5 \]

Step 5: Since \( l, m \) are common to both, find \( \frac{a l}{c l} = \frac{3}{6} = \frac{1}{2} \). So \( \frac{a}{c} = \frac{1}{2} \).

Step 6: From \( b l + a m = -5 \) and \( d l + c m = -1 \), express in terms of \( a, b, c, d, l, m \).

Step 7: From \( a l = 3 \) and \( c l = 6 \), write:
\[ a = \frac{3}{l}, \quad c = \frac{6}{l} \] Similarly, from \( b m = P \) and \( d m = -5 \):
\[ b = \frac{P}{m}, \quad d = \frac{-5}{m} \]

Step 8: Substitute \( a, b, c, d \) into the middle terms equations:
\[ b l + a m = -5 \implies \frac{P}{m} l + \frac{3}{l} m = -5 \] \[ d l + c m = -1 \implies \frac{-5}{m} l + \frac{6}{l} m = -1 \]

Step 9: Multiply first equation by \( m \) and second by \( m \):
\[ P l + 3 m^2 / l = -5 m \] \[ -5 l + 6 m^2 / l = - m \]

Step 10: Rearrange second equation:
\[ 6 \frac{m^2}{l} = -m + 5 l \] Multiply by \( l \):
\[ 6 m^2 = l(-m + 5 l) = -l m + 5 l^2 \] Rearranged:
\[ 6 m^2 + l m - 5 l^2 = 0 \] Consider \( m \) as variable:
\[ 6 m^2 + l m - 5 l^2 = 0 \] Discriminant:
\[ \Delta = l^2 + 4 \times 6 \times 5 l^2 = l^2 + 120 l^2 = 121 l^2 \] So:
\[ m = \frac{-l \pm 11 l}{12} = l \times \frac{-1 \pm 11}{12} \] Possible values:
\[ m = l \times \frac{10}{12} = \frac{5l}{6} \quad \text{or} \quad m = l \times \frac{-12}{12} = -l \]

Step 11: Substitute back into the first equation:
\[ P l + 3 \frac{m^2}{l} = -5 m \] For \( m = \frac{5l}{6} \):
\[ P l + 3 \frac{\left(\frac{5l}{6}\right)^2}{l} = -5 \times \frac{5l}{6} \] \[ P l + 3 \frac{\frac{25 l^2}{36}}{l} = -\frac{25 l}{6} \] \[ P l + 3 \times \frac{25 l}{36} = -\frac{25 l}{6} \] \[ P l + \frac{75 l}{36} = -\frac{25 l}{6} \] Multiply by 36:
\[ 36 P l + 75 l = -150 l \] Divide by \( l \):
\[ 36 P + 75 = -150 \] \[ 36 P = -225 \implies P = -\frac{225}{36} = -\frac{25}{4} \]

For \( m = -l \):
\[ P l + 3 \frac{(-l)^2}{l} = -5 (-l) \] \[ P l + 3 \frac{l^2}{l} = 5 l \] \[ P l + 3 l = 5 l \] \[ P l = 2 l \implies P = 2 \]

Step 12: Sum of all possible values of \( P \):
\[ -\frac{25}{4} + 2 = -\frac{25}{4} + \frac{8}{4} = -\frac{17}{4} \]

Therefore,
\[ \boxed{ -\frac{17}{4} } \]