The normal to the curve `2x^2+y^2=12` at the point `(2,2)` cuts the curve again at (a) `(-(22)/9,-2/9)` (b) `((22)/9,2/9)` `(-2,-2)` (d) none of these

Options

A. `(-22//9, -2//9)`
B. `(22//9,2//9)`
C. `(-2,-2)`
D. none of these

Answer

A

Explanation

We have,
`2x^(2)+y^(2)=12 " "...(i)`
Differentiating w.r.t. x, we get
`4x+2y(dy)/(dx)=0 rArr (dy)/(dx)=-(2x)/(y) rArr ((dy)/(dx))_((2","2))=-2. `
The equation of the normal at (2,2) is
` y-2=(1)/(2)(x-2) rArr x-2y+2 =0 " "...(ii)`
Solving (i) and (ii), we obtain that the coordinates of their points of intersection are (2,2) and `(-22//9,-2//9)`
Hence, the normal to the curve at (2,2) cuts it again at `(-22//9,-2//9). `