Ellipse eccentricity?

Geometry Level 4

If the line 2 p x + y 1 − p 2 = 1 2px+y\sqrt{1-p^{2}}=1 always touches the ellipse x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 ∀ p ∈ ( − 1 , 1 ) − { 0 } \forall \ p \in (-1,1) - \{0\} , find the eccentricity of this ellipse.

3 2 \frac{\sqrt{3}}{2} 7 3 \frac{\sqrt{7}}{3} 1 2 \frac{1}{\sqrt{2}} 7 4 \frac{\sqrt{7}}{4}

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Aryan Goyat
Mar 12, 2016

since the parameter of p is from (-1,+1) so we can take it sin(y) remember y is fixed,

since points in the ellipse can be taken as ( a c o s ( x ) , b s i n ( x ) ) (acos(x),bsin(x))

so putting this points on the line 2 a s i n ( y ) c o s ( x ) + b c o s ( y ) s i n ( x ) = 1 2asin(y)cos(x)+bcos(y)sin(x)=1

now since we want the boundary line condition(ie only one value of xor simply one value of y+x) then s i n ( y + x ) = 1 sin(y+x)=1 this implies 2 a = 1 , b = 1 2a=1,b=1

ecentricity=sqrt(3)/2

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