If the line `2px+ysqrt(5-6p^(2))=1, p in [-sqrt(5)/(6),sqrt(5)/(6)]`, always touches the standard ellipse. Then find the eccentricity of the standard ellipse.

Answer

`1//sqrt(3)`

Explanation

Standard Ellipse : `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Equations of tangent to ellipse having slope m is
`y=mx+-sqrt(a^(2)m^(2)+b^(2))" "(1)`
Given equation of tangent is `2px+ysqrt(5-6p^(2))=1`
Comparing, we get
`m=-(2p)/(sqrt(5-6p^(2)))a^(2)m^(2)+b^(2)=((1)/(sqrt(5-6p^(2))))^(2)`
`rArra^(2)(4p^(2))/((5-6p^(2)))+b^(2)=(1)/(5-6p^(2))`
`rArr4a^(2)p^(2)+b^(2)(5-6p^(2))-1=`
`rArrp^(2)(4a^(2)-6b^(2))+5b^(2)-=0`
Equation (1) should be true for all value of `p in [-sqrt((5)/(6)),sqrt((5)/(6))]`
`:. 4a^(2)=6b^(2)and 5b^(2)-1=0`
`rArra^(2)=(3)/(10) and b^(2)=(1)/(5)`
`rArre=sqrt(1-(10)/(5xx3))=sqrt((5)/(15))=(1)/(sqrt(3))`