If the normal at an end oof a lasrurectum of an ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1` pases through one extremly of the minor axis, show that the eccentricity of the ellipse is given by `e^(4)+e^(2)-1=0` or `e^(2)=sqrt((5)-1/(2))`

Options

A. `e^(4)-e^(2)+1=0`
B. `e^(2)-e+1=0`
C. `e^(2)+e+1=0`
D. `e^(4)+e^(2)-1=0`

Answer

D

Explanation

the coordinates of the extremity of the latusrectum which lies in the first quadant are `(ae,b^(2)//a).`
the equation of the normal ata `(x_(1),y_(1))` is
`(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=(a^(2)-b^(2))`
Therefore the equation of the normal at `(ae,b^(2)//a)` is
`(a^(2)x)/(ae)- (b^(2)y)/(b^(2)//a)=a^(2)-b^(2)`
`implies ax- ae =e (a^(2)-b^(2)) implies ax- aey =ea^(2)e^(2)implies x-ey^(2)implies x-ey =ae^(3)`
this passes through the extremity of the minor axis i.e (0,-b)
` therefore 0+ e-ae^(3)=0`
`implies b=ae^(2)`
`implies b^(2) =a^(2)e^(4)implies a^(2)(1-e^(2))=a^(2)e^(4)implies e^(4)+e^(2)-1=0`