A) \[12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0\]
B) \[12{{x}^{2}}+4{{y}^{2}}-24ax+9{{a}^{2}}=0\]
C) \[12{{x}^{2}}-4{{y}^{2}}+24ax+9{{a}^{2}}=0\]
D) \[12{{x}^{2}}+4{{y}^{2}}+24ax+9{{a}^{2}}=0\]
Correct Answer: A
Solution :
Let\[C\equiv (h,\ k)\], radius \[=r\] Co-ordinates of \[A\equiv \left[ \frac{ah}{a+r},\ \frac{ak}{a+r} \right]\] Co-ordinates of \[B\equiv \left[ \frac{2ar+2ah}{2a+r},\ \frac{2ak}{2a+r} \right]\] Putting co-ordinates of A and B in \[{{S}_{1}},\ {{S}_{2}}\] respectively and eliminating r, we get the locus \[12{{x}^{2}}-4{{y}^{2}}-24ax+9{{a}^{2}}=0\]. Aliter: Since it touches \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]and\[{{x}^{2}}+{{y}^{2}}-4ax=0\], therefore \[r+a=\sqrt{{{h}^{2}}+{{k}^{2}}}\] ?.(i) \[r+2a=\sqrt{{{(h-2a)}^{2}}+{{k}^{2}}}\] ?.(ii) From (i), putting the value of r in (ii), we get \[-a+\sqrt{{{h}^{2}}+{{k}^{2}}}+2a=\sqrt{{{(h-2a)}^{2}}+{{k}^{2}}}\] On simplification, we get the required locus.You need to login to perform this action.
You will be redirected in
3 sec