Plan : As to maximum or minimise area of triangle, we should find area in terms of parametric coordinates and use second derivative test.
Here, tangent at ` P( 2 cos theta, sqrt 3 sin theta )` is
`" " (x)/(2) cos theta + ( y)/(sqrt 3) sin theta = 1 `
` therefore " " R ( 2 sec theta , 0)`
`rArr " " Delta ` = Area of ` Delta PQR`
` " " = (1)/(2) ( 2 sqrt 3 sin theta ) ( 2sectheta - 2 cos theta ) `
` " " = 2 sqrt 3 * sin ^(3) theta // cos theta " "` ... (i)
Since, ` " " (1)/(2) le h le 1 `
` therefore " " (1)/(2) le 2 cos theta le 1 `
`rArr " " (1)/(4) le cos theta le (1)/(2)" " ` ... (ii)
`therefore( d theta )/( d theta ) = (2 sqrt 3 {cos theta * 3 sin ^(2) theta cos theta - sin ^(3) theta ( - sin theta )})/( cos ^(2)theta )`
` " " = ( 2 sqrt 3 * sin ^(2) theta )/( cos^(2) theta ) [ 3cos ^(2) theta + sin ^(2) theta] `
` " " = ( 2 sqrt 3 sin ^(2) theta)/( cos ^(2) theta ) * [ 2 cos ^(2) theta + 1 ]`
` " " = 2 sqrt 3 tan ^(2) theta ( 2 cos ^(2) theta + 1 ) gt 0`
When ` " " (1)/(4) le cos theta le (1)/(2)`
` therefore Delta _1 = Delta _(max)` occurs at `cos theta = (1)/(4) = (( 2sqrt 3 * sin ^(2) theta ) /( cos theta ))`
When ` cos theta = (1)/(4) = ( 45 sqrt 5)/( 8)`
`Delta_2 = Delta_(min) ` occurs at ` cos theta = (1)/(2)`
` " " = (( 2 sqrt 3 sin ^(3) theta )/( cos theta )) `
When ` cos theta = (1)/(2) = (9)/( 2)`
` therefore " " ( 8)/( sqrt 5) Delta_1 - 8 Delta_2 = 45 - 36 =9`
