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 . Locus of the point of intersection of perpendicular tangents to the ellipse is:

Options

A. `x^(2)+y^(2)=125`
B. `x^(2)+y^(2)=150`
C. `x^(2)+y^(2)=200`
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