If the normal at the point `P( theta)` to the ellipse `(x^(2))/(14)+(y^(2))/(5)=1` intersects it again at the point Q `(2 , theta)` then ` cos theta` is equal to

Options

A. `(2)/(3)`
B. `-(2)/(3)`
C. `(3)/(2)`
D. `-(3)/(2)`

Answer

B

Explanation

we have `(x^(2))/(14)+(y^(2))/(5)=1`
the equation of the normal to this ellispe at `p(sqrt(14)cos theta , sqrt(5) sin theta)` is
`( sqrt(14)cos theta )x-(sqrt(5)"cosec" theta) y=9`
this meets the ellipse (i) at Q `(sqrt(14) cos 2 , theta , sqrt(5) sin 2, theta )`
`therefore 14 sec theta cos 2 theta - 5 "cosec"theta sin 2 theta =9`
`implies 18 cos^(2) theta -9 cos theta -14 =0`
`implies ( 6 cos 0-7)(3 cos theta +2)-0`
`implies 3 cos theta +2=0implies cos theta =-2//3.`