about to only mathematics

Options

A. 9
B. 13
C. 4
D. 5

Explanation

Let the P be ` (a cos theta b, sin theta)`
The equation of the tangent at point P is `(x)/(a)cos theta+(y)/(b)sin theta=1`
The point Q is `(0,b s "cosec" theta)`
The quation of chord A'P is
`y-0=(b sin theta)/(a cos theta+a)(x+a)`
Putting x=0, we have
`y=(b sin theta)/(cos thea+1)`
Then,
`OQ^(2)-MQ^(2)=b^(2)"coses"^(2)theta-(b "cosec" theta-(b sin theta)/(cos theta+1))^(2)`
`=(2b^(2))/(costheta+1)-(b^(2)sin theta)/((cos theta+1)^(2))`
`=(b^(2))/(cos theta+1)(b^(2)sin theta)/((cos theta+1)^(2))`
`=(b^(2))/(cos theta+1)((2 cos theta+2-sin^(2)theta)/(cos theta+1))`
`=(b^(2))/(cos theta+1)((2 cos theta+1+cos^(2)theta)/(cos theta+1)) =b^(2)=4`