A) \[2x-y\pm 5=0\]
B) \[2x-y+5=0\]
C) \[2x-y-5=0\]
D) \[2x+y+5=0\]
Correct Answer: A
Solution :
Let the equation of line be\[y=mx+c\]. Since this line is tangent to the circle \[{{x}^{2}}+{{y}^{2}}=5\] \[\therefore \] \[c=\pm a\sqrt{1+{{m}^{2}}}\] \[=\pm \sqrt{5}\sqrt{1+{{m}^{2}}}\] ...(i) Also, the above line is tangent to the parabola \[{{y}^{2}}=40x\] \[\therefore \] \[c=\frac{a}{m}=\frac{10}{m}\] ?..(ii) From Eqs, (i) and (ii), \[\Rightarrow \] \[\frac{10}{m}=\pm \sqrt{5}\sqrt{1+{{m}^{2}}}\] \[\Rightarrow \] \[\frac{20}{{{m}^{2}}}=1+{{m}^{2}}\] \[\Rightarrow \] \[{{m}^{4}}+{{m}^{2}}-20=0\] \[\Rightarrow \] \[({{m}^{2}}+5)({{m}^{2}}-4)=0\] \[\Rightarrow \] \[{{m}^{2}}=4,{{m}^{2}}\ne -5\] \[\Rightarrow \] \[m=\pm 2\] \[\Rightarrow \] \[c=\pm 5\] \[\therefore \] \[y=\pm 2x\pm 5\]
You need to login to perform this action.
You will be redirected in
3 sec
