Let ABC be an equilateral triangle inscribed in the circle `x^2+y^2=a^2` . Suppose pendiculars from A, B, C to the ellipse `x^2/a^2+y^2/b^2=1,(a > b)` meets the ellipse respectivelily at P, Q, R so that P, Q , R lies on same side of major axis as A, B, C respectively. Prove that the normals to the ellipse drawn at the points P Q nad R are concurrent.

Explanation

Let A , B and C, be the points on the cicle whose cooridnates are
`A(a cos theta, a sin theta)`
`B(a cos (theta+(2pi)/(3)),asin(theta+(2pi)/(3)))`
`C(a cos (theta+(4pi)/(3)),asin(theta+(4pi)/(3)))`
Hence, `P-=(a cos theta, b sin theta)`
`Q-=(a cos (theta+(2pi)/(3)),asin(theta+(2pi)/(3)))`
`R-=(a cos (theta+(4pi)/(3)),asin(theta+(4pi)/(3)))`
It is given that P,Q and R are on the same side of the x-axis as A,B, and C.
So, the required nomrmals ot the ellipse at P,Q and R are
`ax sec theta-"by cosec" theta=a^(2)-b^(2)" "(1)`
`ax sec(theta+(2pi)/(3)-"by cosec"(theta+(2pi)/(3))=a^(2)-n^(2)" "(2)`
`ax sec(theta+(4pi)/(3)-"by cosec"(theta+(4pi)/(3))=a^(2)-n^(2)" "(3)`

Now, `Delta`= `|{:(sec theta,"cosec"theta,1),(sec (theta+(2pi)/(3)),"cosec"(theta+(2pi)/(3)),1),(sec (theta+(2pi)/(3)),"cosec"(theta+(2pi)/(3)),1):}|`
Multiplying `R_(1),R_(2) and R_(3)`, by `sin theta cos, theta, sin (theta+(2pi)/(3))co (theta+(2pi)/(3))and sin (theta+(4pi)/(3))` respectively, we get
`Delta=(1)/(k)|{:(sin theta,cos theta,sin2theta),(sin (theta+(2pi)/(3)),cos(theta+(2pi)/(3)),sin (2 theta+(4pi)/(3))),(sin (theta-(2pi)/(3)),cos(theta-(2pi)/(3)),sin (2 theta-(4pi)/(3))):}|`
Where
`k=2sin theta cos theta sin(theta+(2pi)/(3))cos(theta-(2pi)/(3))sin(theta+(4pi)/(3))cos (theta+(4pi)/(3))`
Operating `R_(2)toR_(2)+R_(3)`
`Delta=(1)/(k)|{:(sin theta,cos theta,sin2theta),(2sin theta*cos.(2pi)/(3),2cos theta*cos.(2pi)/(3),2sin theta*cos.(4pi)/(3)),(sin(theta-(2pi)/(3)),cos(theta-(2pi)/(3)),sin (2theta-(4pi)/(3))):}|`
`=(1)/(k)|{:(sin theta,cos theta,sin2theta),(-sin theta ,-cos theta,-sin theta),(sin(theta-(2pi)/(3)),cos(theta-(2pi)/(3)),sin(2theta-(4pi)/(3))):}|`