A) \[\frac{a-c}{b}>2\]
B) \[\frac{b}{a-c}>2\]
C) \[\frac{b}{a+c}>2\]
D) None of these
Correct Answer: B
Solution :
Equation to any normal is \[y=mx-2am-a{{m}^{3}}\] Equation to any normal is \[y=m(x-b)-2cm-c{{m}^{3}}\] If there is any common normal, then they must be identical. As the coefficients of\[x\]and y are equal so the constant terms will be also equal, hence \[-2am-a{{m}^{3}}=-bm-2mc-c{{m}^{3}}\] or \[m[{{m}^{2}}(c-a)-2a+b+2c]=0\] So, either\[m=0\]or \[{{m}^{2}}=\frac{2a-b-2c}{c-a}\] Then, \[m=\sqrt{\left( \frac{2(a-c)-b}{c-a} \right)}=\sqrt{\left( -2-\frac{b}{c-a} \right)}\] If the value of m is real and not zero, then \[-2-\frac{b}{c-a}>0,\frac{-b}{c-a}>0\]or \[\frac{b}{a-c}>2\]
You need to login to perform this action.
You will be redirected in
3 sec
