Question:
If the point \((p,5)\) lies on the line parallel to the \(y\)-axis and passing through the intersection of the lines \(2\left(a^{2}+1\right)x+by+4\left(a^{3}+a\right)=0\) and \(\left(a^{2}+1\right)x-3by+2\left(a^{3}+a\right)=0\), then \(p\) is equal to :
If the point \((p,5)\) lies on the line parallel to the \(y\)-axis and passing through the intersection of the lines \(2\left(a^{2}+1\right)x+by+4\left(a^{3}+a\right)=0\) and \(\left(a^{2}+1\right)x-3by+2\left(a^{3}+a\right)=0\), then \(p\) is equal to :
Show Hint
A line parallel to the \(y\)-axis is \(x=\text{constant}\); so \(p\) is just the \(x\)-coordinate of the intersection point. Eliminate \(y\).
Updated On: Nov 12, 2026
- \(-2a\)
- \(-3a\)
- \(2a\)
- \(3a\)
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The Correct Option is A
Solution and Explanation
Put \(A=a^{2}+1\), so \(a^{3}+a=aA\). The lines are \(2Ax+by+4aA=0\) and \(Ax-3by+2aA=0\).
Multiply the first by 3 and add to the second: \(6Ax+3by+12aA+Ax-3by+2aA=0\), i.e. \(7Ax+14aA=0\).
Since \(A\neq 0\), \(x=-2a\).
The line through the intersection parallel to the \(y\)-axis is \(x=-2a\); it contains \((p,5)\), so \(p=-2a\).
Answer: (1) \(-2a\)
Multiply the first by 3 and add to the second: \(6Ax+3by+12aA+Ax-3by+2aA=0\), i.e. \(7Ax+14aA=0\).
Since \(A\neq 0\), \(x=-2a\).
The line through the intersection parallel to the \(y\)-axis is \(x=-2a\); it contains \((p,5)\), so \(p=-2a\).
Answer: (1) \(-2a\)
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