The focal chord to `y^2=16 x` is tangent to `(x-6)^2+y^2=2.` Then the possible value of the slope of this chord is (a)`{-1,1}` (b) `{-2,2}` (c)`{-2,1/2}` (d) `{2,-1/2}`

Answer

1

Explanation

(1) The focus of `y^(2)=16x` is (4,0).
Any focal chord is
y-0=m(x-4)
or mx-y-4m=0
This the focal chord touches the circle `(x-6)^(2)+y^(2)=2`.
Then the distance from the center of the circle to this chord is equal to the radius of the circle, i.e.,
`(|6m-4m|)/(sqrt(m^(2)+1))=sqrt(2)`
`or" "2m=sqrt(2)*sqrt(m^(2)+1)`
`or" "2m^(2)=m^(2)+1orm^(2)=1`
`:." "m=pm1`