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Two straight lines (y-b)=m1(x+a) and (y-...

Two straight lines `(y-b)=m_1(x+a)` and `(y-b)=m_2(x+a)` are the tangents of `y^2=4a xdot` Prove `m_1m_2=-1.`

Text Solution

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Clearly, both the lines pass through (-a,b), which is a point lying on the directrix of the parabola.
Thus, `m_(1)m_(2)=-1`, because tangents drawn from any point on the directrix are always mutually perpendicular.
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Knowledge Check

  • If the straight line y=my+1 be the tangent of the parabola y^2=4x at the point (1,2), then the value of m will be

    A
    1
    B
    2
    C
    -1
    D
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  • The condition for which the striaght lines l_(1)x+m_(1)y+n_(1)=0andl_(2)x+m_(2)y+n_(2)=0 are perpendicular to each other is -

    A
    `l_(1)l_(2)+m_(1)m_(2)=0`
    B
    `l_(1)m_(1)+l_(2)m_(2)=0`
    C
    `l_(1)m_(2)+l_(2)m_(1)=0`
    D
    `l_(1)l_(2)-m_(1)m_(2)=0`

  • The lines x+y-1=0,(m-1)x+(m^(2)-7)y-5=0and(m-2)x+(2m-5)y=0 are -

    A
    concurrent for three values of m
    B
    concurrent for one value of m
    C
    concurrent for no value of m
    D
    parallel for m=3

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