The area of the rectangle formed by the perpendicular from the center of the standard ellipse to the tangent and normal at its point whose eccentric angle is `pi/4`, is

Options

A. `((a^(2) - b^(2))/(a^(2) + b^(2)))ab`
B. `((a^(2) + b^(2))/(a^(2) - b^(2)))ab`
C. `(a^(2) - b^(2))/(a^(2) + b^(2))`
D. `(a^(2) + b^(2))/a^(2) - b^(2))`

Answer

A

Explanation

The given point is `(a cos pi//4, b sin pi//4)` i.e. `(a//sqrt2, b//sqrt2).`
So, the equation of the tangent at this point is
`x/a + y/b = sqrt2" "…(i)`
`therefore p_(1) =` Length of the perpendicular form (0,0) on line (i)
`rArr p_(1) = |(0/a + 0/b - sqrt2)/(sqrt(1//a^(2) + 1//b^(2)))| = (sqrt(2) ab)/(sqrta^(2) + b^(2))`
Equation of the normal at `(a/sqrt2, b/sqrt2)` is
`(a^(2)x)/(a//sqrt2) - (b^(2)y)/(b//sqrt2) = a^(2) - b^(2) rArr sqrt(2) ax - sqrt2 ax - sqrt2 by = a^(2) - b^(2) ...(ii)`
`therefore p_(2) =` Length of the perpendicular form (0, 0) on line (ii)
`rArr p_(2) = (a^(2) - b^(2))/(sqrt((sqrt(2a))^(2) + ( - sqrt(2b))^(2)) ) = (a^(2) - b^(2))/(sqrt(2(a^(2) + b^(2)))) `
`therefore` Area of the rectangle
`rArr p_(1)p_(2) = (sqrt2 ab)/(sqrt(a^(2) + b^(2))) xx (a^(2) - b^(2))/(sqrt(2(a^(2) + b^(2)))) = ((a^(2) - b^(2))/(a^(2) + b^(2)))ab.`