Find the minimum value of `(2-a-4 sec theta)^(2)+(a-3 tan theta)^(2), a in R.`

Explanation

We have `f(theta)=(2-a-4 sec theta)^(2)+(a-3tan theta)^(2)`
This is square of the distance between the points `P(2-a,a)` and Q`(4 sec theta, 3 tan theta)`.
Point P lies on the line
`x+y-2=0" (1)"`
and Q lies on hyperbola
`(x^(2))/(16)-(y^(2))/(9)=1`
So, minimum value of `f(theta)` is square of the distnace between the line (l) and the tanent to hyperbola at point P which is parallel to the line (l).
Equations of tangents to `(x^(2))/(16)-(y^(2))/(9)=1` having slope `-1` ar e given by
`y=-xpmsqrt7`
`" or"x+y+sqrt7=0" (2)"`
`"and "x+y-sqrt7=0`
Distance between lines (1) and (2) is `(sqrt7+2)/(sqrt2)`.
Distance between lines (1) and (3) is `(sqrt7-2)/(sqrt2)`
So, required minimum distance is `(sqrt7-2)/(sqrt2).`
`therefore" "f(theta)=((sqrt7-2)/(sqrt2))^(2)`