InterviewSolution
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InterviewSolution
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An ellipse has eccentricity `1/2` and one focus at the point `P(1/2,1)`. Its one directrix is the comionand tangent nearer to the point the P to the hyperbolaof `x^2-y^2=1` and the circle `x^2+y^2=1`.Find the equation of the ellipse. |
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Answer» Correct Answer - A::B::C there are two common tangent to the circle `x^(2)+y^(2)=1` and the hyperbola `x^(2)-y^(2)=1` these are x=1 and x=-1. But x=1 is nearer to the point P(1/2,1). Now, if Q(x,y) is any point on the eillpse then its distance from the foruc is `QP= sqrt((x-1//2)^(2)+(y-1)^(2))` and its distance from the directrix is |x-1|. By definition of ellipse , `QP=e|x-1|implies sqrt((x-(1)/(2))^(2)+(y+1)^(2))=(1)/(2)|x-1|` `implies (x-(1)/(2))^(2)+(y-1)^(2)=(1)/(4)(x-1)^(2)` `implies x^(2)-x+(1)/(4)+y^(2)-2y+1=(1)/(4)(x^(2)-2x+1)` `implies 4x^(2)-4x+1+4y^(2)-8y+4=x^(2)-2x+1` `implies 3x^(2)-2x+4y^(2)-8y+4=0` `3[(x-(1)/(3))^(2)-(1)/(9)]+4(y-1)^(2)=0` `implies 3(x-(1)/(3))^(2)+4(y-1)^(2)=(1)/(3)` `implies ((x-(1)/(3)^(2)))/(1//9)`+((y-1)^(2))/(1//12)=1` |
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