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If the curves `(x^2)/4+y^2=1`and `(x^2)/(a^2)+y^2=1`for a suitable value of `a`cut on four concyclic points, the equation of the circle passingthrough these four points is`x^2+y^2=2`(b) `x^2+y^2=1``x^2+y^2=4`(d) none of theseA. `x^(2)+y^(2)=2`B. `x^(2)+y^(2)=1`C. `x^(2)+y^(2)=4`D. none of these |
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Answer» The equation of conic through the point of intersection of given two ellipse is `((x^(2))/(4)+y^(2)-1)+lambda((x^(2))/(a^(2))+y^(2)-1)=0` or `x^(2)((1)/(4)+(lambda)/(a^(2)))+y^(2)(1+lamda)=1+lambda` or `x^(2)((a^(2))/(4a^(2)(1+lambda)))+y^(2)=1` Therefore , the circle is `x^(2)+y^(2)=1` |
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