A) \[3(x+y)+4=0\]
B) \[8(2x+y)+3=0\]
C) \[4(x+y)+3=0\]
D) \[x+2y+3=0\]
Correct Answer: C
Solution :
Origin \[(0,0)\] is the only point common to x-axis and y-axis. \[\Rightarrow \] Origin \[(0,0)\] is the common vertex Let the equation of 2 parabola be \[{{y}^{2}}=4ax\] and \[{{x}^{2}}=4by\] Latus rectum\[=3\] \[\Rightarrow 4a=4b=3\] \[\Rightarrow a=b=\frac{3}{4}\] \[\therefore \] The 2 parabolas are \[{{y}^{2}}=3x\] and\[{{x}^{2}}=3y\] Let \[y=mx+c\] be the common tangent \[{{y}^{2}}=3x\] \[\Rightarrow {{(mx+c)}^{2}}=3x\] \[\Rightarrow {{m}^{2}}{{x}^{2}}+(2mc-3)x+{{c}^{2}}=0\] The tangent touches at only one point \[\Rightarrow {{b}^{2}}-4ac=0\] \[\Rightarrow {{(2mc-3)}^{2}}-4{{m}^{2}}{{c}^{2}}=0\] \[\Rightarrow 4{{m}^{2}}{{c}^{2}}+9-12mc-4{{m}^{2}}{{c}^{2}}=0\] \[\Rightarrow c=\frac{9}{12m}=\frac{3}{4m}........(1)\] \[{{m}^{2}}=-c=\frac{-3}{4m}\] \[{{x}^{2}}=3y\] \[\Rightarrow {{x}^{2}}=3(mx+c)\] \[\Rightarrow {{x}^{2}}-3mx-3c=0\] Tangent touches at only one point \[\Rightarrow {{b}^{2}}-4ac=0\] \[\Rightarrow 9{{m}^{2}}-4(1)(-3c)=0\] \[\Rightarrow 9{{m}^{2}}=-12c...........(2)\] From (1) and (2) \[{{m}^{2}}=\frac{-4c}{3}=\frac{-4}{3}\left( \frac{3}{4m} \right)\] \[\Rightarrow {{m}^{3}}=-1\] \[\Rightarrow m=-1\] \[\Rightarrow c=\frac{-3}{4}\] \[\therefore y=mx+c=-x-\frac{3}{4}\] \[\Rightarrow 4(x+y)+3=0\]
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