The focal chord of `y^(2)=16x` is tangent to `(x-6)^(2)+y^(2)=2`. Then the possible value of the square of slope of this chord is __________ .
The focal chord of `y^(2)=16x` is tangent to `(x-6)^(2)+y^(2)=2`.
Then the possible value of the square of slope of this chord is __________ .
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Correct Answer - 1
(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`
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