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How to find the equation of tangents drawn to a parabola `y^2 = 4x` from a point (-8,0) ?

Answer» Equation of tangents of a parabola is given by,
`y = mx+a/m->(1)`
In the given parabola, `a =1`
So, equations of tangents will be,
`y = mx+1/m`
Now, tangents are passing through `(-8,0)`.
`:. 0 = -8m+1/m`
`=>8m^2 = 1=> m = +-1/(2sqrt2)`
So, equation otangents will be,
For `m = 1/(2sqrt2)`
`y = x/(2sqrt2)+2sqrt2`
For `m = -1/(2sqrt2)`
`y = -x/(2sqrt2)-2sqrt2`


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