InterviewSolution
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InterviewSolution
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For all real p, the line 2px+ysqrt(1-p^(2))=1 touches afixed ellipse whose axex are the coordinate axes The locus of the point of intersection of perpendicular tangent is |
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Answer» `x^(2)+y^(2)=5//4` `(x^(2))/(y^(2))+(y^(2))/(b^(2))=1` The line `y==mx+-sqrt(a^(2)m^(2)+b^(2))`touches, the ellipse for all m. Hence, it is identical with `y=-(2X)/(sqrt(1-p^(2)))+(1)/(sqrt(1-p^(2)))` Hence, `m=-(2p)/(sqrt(1-p^(2)))` and `a^(2)m^(2)+b^(2)=(1)/(1-p^(2))` or `a^(2)(4p^(2))/(a-p^(2))+b^(2)(1)/(1-p^(2))` or `p^(2)(4A^(2)-b^(2))+b^(2)-1=0` This equation is true for all real p if `b^(2)=1 and 4a^(2)=b^(2)` `b^(2)=1 and a^(2)=(1)/(4)` Therefore, the equation of the ellipse is `(x^(2))/(1//4)+(y^(2))/(1)=1` If e is its ECCENTRICITY, then `(1)/(4)=1-e^(2) or e^(2)=(3)/(4) or e=(sqrt(3))/(2)` be `=(sqrt(3))/(2)` Hence, the FOCI are `(0,+-sqrt(3)//2)` The equation of director circle is `x^(2)+y^(2)=(5)/(4)` |
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