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JEE MainMathematicsModerateStraight Lines

If the point (p, 5) lies on the line parallel to the y-axis and
passing through the intersection of the lines
2(a2+1)x+by+4(a3+a)=02\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0 and
(a2+1)x−6by+2(a3+a)=0\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0 then p
is equal to:
ij fLFkr gS tks y- lekarj gS rFkk
2(a2+1)x+by+4(a3+a)=02\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0 rFkk
(a2+1)x−6by+2(a3+a)=0\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0

Show answer and solution
Line parallel to y-axis is x = p
y- lekUrj js[kk x = p
2(a2+1)p+by+4(a3+a)=0         …(i)2\left( a^{2} + 1 \right)p + by + 4\left( a^{3} + a \right) = 0\ \ \ \ \ \ \ \ \ \ldots(i)
(a2+1)p−3by+2(a3+a)=0         …(ii)\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0\ \ \ \ \ \ \ \ \ \ldots(ii)
Equation (i) is (lehdj.k (i) gS )
6(a2+1)p+3by+12(a3+a)=06\left( a^{2} + 1 \right)p + 3by + 12\left( a^{3} + a \right) = 0
(a2+1)p−3by+2(a3+a)=0\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0
Adding tksM+us ij
7(a2+1)p+14(a3+a)=07\left( a^{2} + 1 \right)p + 14\left( a^{3} + a \right) = 0
7(a2+1)p=−14(1)(a2+1)⇒p=−2a7\left( a^{2} + 1 \right)p = - 14(1)\left( a^{2} + 1 \right) \Rightarrow p = - 2a

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

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