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Tangent are drawn from the point (-1,2) ...

Tangent are drawn from the point `(-1,2)` on the parabola `y^2=4x` . Find the length that these tangents will intercept on the line `x=2.`

Text Solution

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The equation of the pair of tangents from point `(x_(1),y_(1))` is `SS_(1)=T^(2)`
`or(y^(2)-4x)(y_(1)^(2)-4x_(1))={yy_(1)-2(x+x_(1))}^(2)`
Then equation of the pair of tangents from (-1,2) is
`(y^(2)-4x)(4+4)=[2y-2(x-1)]^(2)=4(y-x+1)^(2)`
`or2(y^(2)-4x)=(y-x+1)^(2)`
Solving with the x=2, we get
`2(y^(2)-8)=(y-1)^(2)`
`ory^(2)+2y-17=0`
`"where "y_(1)+y_(2)=-2andy_(1)y_(2)=-17`
Now, `|y_(1)-y_(2)|^(2)=(y_(1)+y_(2))^(2)-4y_(1)y_(2)=4-4(-17)=72`
`:." "|y_(1)-y_(2)|=sqrt(72)=6sqrt(2)`
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