A) \[\frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1\]
B) \[\frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y+1)}^{2}}}{1/12}=1\]
C) \[\frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y-1)}^{2}}}{1/12}=1\]
D) \[\frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y+1)}^{2}}}{1/12}=1\]
Correct Answer: A
Solution :
There are two common tangents to the circle \[{{x}^{2}}+{{y}^{2}}=1\] and the hyperbola \[{{x}^{2}}-{{y}^{2}}=1.\] These are \[x=1\] and \[x=-1\] Out of these, \[x=1\] is nearer to the point \[P(1/2,1)\]. Thus a directrix of the required ellipse is \[x=1.\] If \[Q(x,y)\] is any point on the ellipse, then its distance from the focus is \[QP=\sqrt{{{\left( x-\frac{1}{2} \right)}^{2}}+{{(y-1)}^{2}}}\] and its distance from the directrix \[x=1\]is \[|x-1|\]. By definition of ellipse, \[QP=e|x-1|\] \[\Rightarrow \sqrt{{{\left( x-\frac{1}{2} \right)}^{2}}+{{(y-1)}^{2}}}=\frac{1}{2}|x-1|\] Þ\[3{{x}^{2}}-2x+4{{y}^{2}}-8y+4=0\]or\[\frac{{{\left( x-\frac{1}{3} \right)}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1\].You need to login to perform this action.
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