Let `F_(1)(x_(1),0) and F_(2)(x_(2),0)` for `x_(1)lt0 and x_(2)gt0`, be the foci of the ellipse `(x^(2))/(9)+(y^(2))/(8)=1` . Suppose a parbola having vertex at the origin and focus at `R_(2)` intersects the ellipse at point lt in the first quadrant and point N in the fourth quadrant
If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of quadrilateral `MF_(1)NF_(2)` is

Options

A. `3:4`
B. `4:5`
C. `5:8`
D. `2:3`

Answer

C

Explanation

Equation of tangent at `M(3//2,sqrt(6)) " to " (x^(2))/(9)+(y^(2))/(8)=1 `is
`(3)/(2).(x)/(9)+sqrt(6)(y)/(8)=1`. . . .(i)
which intersect X- axis at (6,0) .
Also , equation of tangent Ata `N(3//2,-sqrt(6))` is
`(3)/(2).(x)/(9)-sqrt(6)= (-sqrt(6))/(2)(x-(3)/(2))`
on solving with y=0, we get Q(7//2,0)

` therefore ` Aera of `Delta MQR=(1)/(2) (6-(7)/(2))sqrt(6)=(5sqrt(6))/(1)` sq units and area of quadrillateral `MF_(1)NF_(2)=2xx(1)/(2){1-(-1)} sqrt(6)`
`= 2 sqrt(6)`sq untis
`therefore ("Area of "Delta MQR)/(" Area of Quadrilateral " MF_(1)NF_(2))=(5)/(8)`