Question:
Suppose the pairs of straight lines
\[
2x^2+axy+3y^2=0
\]
and
\[
2x^2+bxy-3y^2=0
\]
are such that they have one common line with the other two remaining perpendicular. Then the values of \(a\) and \(b\) respectively are:
Suppose the pairs of straight lines
\[
2x^2+axy+3y^2=0
\]
and
\[
2x^2+bxy-3y^2=0
\]
are such that they have one common line with the other two remaining perpendicular. Then the values of \(a\) and \(b\) respectively are:
Show Hint
For a homogeneous second-degree equation representing a pair of lines through the origin, put \(y=mx\) to get the quadratic equation in slopes.
Updated On: Jun 24, 2026
- \(-5,1\)
- \(5,-1\)
- \(5,1\)
- \(5,\dfrac{1}{5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Convert the pair of lines into equations in slope.
Let a line through the origin be \[ y=mx \] For \[ 2x^2+axy+3y^2=0, \] put \(y=mx\).
Then, \[ 2x^2+ax(mx)+3m^2x^2=0 \] \[ x^2(2+am+3m^2)=0 \] Hence, the slopes of the two lines satisfy \[ 3m^2+am+2=0 \] For \[ 2x^2+bxy-3y^2=0, \] put \(y=mx\).
Then, \[ 2x^2+bx(mx)-3m^2x^2=0 \] \[ x^2(2+bm-3m^2)=0 \] Hence, the slopes of the two lines satisfy \[ 3m^2-bm-2=0 \]
Step 2: Use the common line condition.
The two pairs have one common line, so the two quadratic equations in \(m\) have one common root.
Check option (3): \[ a=5,\quad b=1 \] Then the first slope equation becomes \[ 3m^2+5m+2=0 \] Factorizing, \[ 3m^2+5m+2=(3m+2)(m+1)=0 \] So, \[ m=-\frac{2}{3},\quad m=-1 \] The second slope equation becomes \[ 3m^2-m-2=0 \] Factorizing, \[ 3m^2-m-2=(3m+2)(m-1)=0 \] So, \[ m=-\frac{2}{3},\quad m=1 \] Thus, the common slope is \[ m=-\frac{2}{3} \]
Step 3: Check perpendicularity of the remaining lines.
The remaining slopes are \[ -1 \] and \[ 1 \] Their product is \[ (-1)(1)=-1 \] Therefore, the remaining two lines are perpendicular.
Step 4: Final conclusion.
Hence, the required values are \[ \boxed{a=5,\ b=1} \]
Let a line through the origin be \[ y=mx \] For \[ 2x^2+axy+3y^2=0, \] put \(y=mx\).
Then, \[ 2x^2+ax(mx)+3m^2x^2=0 \] \[ x^2(2+am+3m^2)=0 \] Hence, the slopes of the two lines satisfy \[ 3m^2+am+2=0 \] For \[ 2x^2+bxy-3y^2=0, \] put \(y=mx\).
Then, \[ 2x^2+bx(mx)-3m^2x^2=0 \] \[ x^2(2+bm-3m^2)=0 \] Hence, the slopes of the two lines satisfy \[ 3m^2-bm-2=0 \]
Step 2: Use the common line condition.
The two pairs have one common line, so the two quadratic equations in \(m\) have one common root.
Check option (3): \[ a=5,\quad b=1 \] Then the first slope equation becomes \[ 3m^2+5m+2=0 \] Factorizing, \[ 3m^2+5m+2=(3m+2)(m+1)=0 \] So, \[ m=-\frac{2}{3},\quad m=-1 \] The second slope equation becomes \[ 3m^2-m-2=0 \] Factorizing, \[ 3m^2-m-2=(3m+2)(m-1)=0 \] So, \[ m=-\frac{2}{3},\quad m=1 \] Thus, the common slope is \[ m=-\frac{2}{3} \]
Step 3: Check perpendicularity of the remaining lines.
The remaining slopes are \[ -1 \] and \[ 1 \] Their product is \[ (-1)(1)=-1 \] Therefore, the remaining two lines are perpendicular.
Step 4: Final conclusion.
Hence, the required values are \[ \boxed{a=5,\ b=1} \]
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