Find the value of m, for which the line `y=mx + 25sqrt3/3` is a normal to the conic ` x^2 /16 - y^2 / 9 =1`.

Answer

`m=-(2)/(sqrt3)`

Explanation

Equation of normal to `(x^(2))/(16)-(y^(2))/(9)=1` at point `P(4 sec theta, 3 tan theta)` on it is given by
`(4x)/(sec theta)+(3y)/(tan theta)=16+9`
`"or "4x cos theta+3y cot theta=25`
This is of the form `-mx+y=(25sqrt3)/(3)" (given)"`
Comparing rations of coefficients, we get
`(4 cos theta)/(-m)=(3 cot theta)/(1)=sqrt3`
`therefore" "cot theta=(1)/(sqrt3)`
`therefore" "m=-(4 cos theta)/(sqrt3)=-(4)/(sqrt3)(1)/(2)=-(2)/(sqrt3)`