JEE Main 2024 — Mathematics Coordinate Geometry

Let, $P\equiv ({a}^{2},a+1)$P≡(a2,a+1), ${L}_{1}:3x-y+1=0$L1​:3x−y+1=0 and ${L}_{2}:x+2y-5=0$L2​:x+2y−5=0.

Now, origin and $P$P lies on the same side with respect to line ${L}_{1}$L1​.

$$\Rightarrow {L}_{1}(0)\cdot {L}_{1}(P)>0$$⇒L1​(0)⋅L1​(P)>0

$$\Rightarrow 3({a}^{2})-(a+1)+1>0$$⇒3(a2)−(a+1)+1>0

$$\Rightarrow 3{a}^{2}-a>0$$⇒3a2−a>0

$$\Rightarrow a(3a-1)>0$$⇒a(3a−1)>0

$$\Rightarrow a\in (-\infty ,0)\cup (\frac{1}{3},\infty )...(i)$$⇒a∈(−∞,0)∪(31​,∞)...(i)

Now, ${L}_{2}:x+2y-5=0$L2​:x+2y−5=0

Origin and $P$P lies on same side with respect to ${L}_{2}$L2​

$$\Rightarrow {L}_{2}(0)\cdot {L}_{2}(P)>0$$⇒L2​(0)⋅L2​(P)>0

$$\Rightarrow {a}^{2}+2(a+1)-5<0$$⇒a2+2(a+1)−5<0

$$\Rightarrow {a}^{2}+2a-3<0$$⇒a2+2a−3<0

$$\Rightarrow (a+3)(a-1)<0$$⇒(a+3)(a−1)<0

$$\therefore a\in (-3,1)...(\mathrm{ii})$$∴a∈(−3,1)...(ii)

Using intersection of $(i)\text{and}(ii)$(i)and(ii),

$$\Rightarrow a\in (-3,0)\cup (\frac{1}{3},1)$$⇒a∈(−3,0)∪(31​,1)