show that the area of the triangle inscribed in the circle `x^(2)/a^(2)+y^(2)/b^(2)=1` meet the ellipse respictively at P,Q,R so that P,Q,R lie on the same side of the major axis as A,B,C respictively. Prove that the normal to the ellipse drawn at the points P,Q and 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))):}|`