Consider the ellipse whose major and minor axes are x-axis and y-axis, respectively. If `phi` is the angle between the CP and the normal at point P on the ellipse, and the greatest value `tan phi` is `3/4` (where C is the centre of the ellipse). Also semi-major axis is 10 units . A rectangle is inscribed in a ellipse whose sides are parallel to the coordinate axes, then the maximum area of rectangle is

Options

A. 50 units
B. 100 units
C. 25 units
D. none of these

Explanation

Let P be any point on the ellipse t `(a cos theta, b sin theta)`
Therefore, the equation of CP is
`y=((b)/(a) tan theta)x`
The normal to the ellipse at P is
`(ax)/(cos theta)-(bx)/(sin theta)=a^(2)-b^(2)`
Slopes of the lines CPand the normal GP are (b/a) `tan theta` and (a/b) `tan, theta` respectively. Therefore,
`tan phi=((a)/(b) tan theta-(b)/(a)tan theta)/(1+(a)/(b)tan theta(b)/(a)tan theta)`
`(a^(2)-b^(2))/(ab)=(tan theta)/(sec^(2)theta)`
`=(a^(2)-b^(2))/(ab) sin theta cos theta=(a^(2)-b^(2))/(2ab) sin 2 theta`
Therefore , the greatest value of `tan phi` is
`(x^(2)-b^(2))/(2ab)xx1=(a^(2)-b^(2))/(2ab)`
Given that
`(a^(2)-b^(2))/(2ab)=(3)/(2)`
Let `(a)/(b)=t`
`:. t-(1)/(t)=(3)/(2)`
or `2t^(2)-3t-2=0`
` or 2t^(2)-4t+t-2=0`
`or (2t+1)(t-2)`
`(a)/(b)=2`
or `e^(2)=1-(1)/(4)`
or `e=(sqrt(3))/(2`
The rectangle inscribed in the ellipse, whose one vertexis `a(cos theta,b sin theta), "is" (2a cos theta b sin theta)=2ab sin (2theta)` which has maximum, value 2ab. Given that a=10. Then b=5. Therefore, the maximum area is 100