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An ellipse intersects the hyperbola `2x^2-2y =1` orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (b) the foci of ellipse are `(+-1, 0)` (a) equation of ellipse is `x^2+ 2y^2 =2` (d) the foci of ellipse are `(t 2, 0)` (c) equation of ellipse is `(x^2 2y)`A. the equation of the ellipse is `x^(2)+2y^(2)=1`B. the foci of the ellipse are `(pm1,0)`C. the equation of the ellipse is `x^(2)+2y^(2)=4`D. the foci of the ellipse are `(pmsqrt2,0)` |
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Answer» Correct Answer - A::B Let the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`. Since ellipse intersects hyperbola orthogonally. The ellipse and hyperbola will be confocal. So, comparing coordinates of foci `(pma+(1)/(sqrt2),0)=(pm1,0)` `"or "a=sqrt2 and e=(1)/(sqrt2)` Now, `b^(2)=a^(2)(1-e^(2))` `"or "b^(2)=1` Therefore, the equation of the ellipse is `(x^(2))/(2)+(y^(2))/(1)=1` |
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