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.`

If the line (y-b)=m_(1)(x+a) and (y-b)=m_(2)(x+a) are the tangents to the parabola y^(2)=4ax then

If y+b=m_(1)(x+a) and y+b=m_(2)(x+a) are two tangents to y^(2)=4ax then m_(1)times m_(2)=

If y+b=m_(1)(x+a) and y+b=m_(2)(x+a) are two tangents to the paraabola y^(2)=4ax then

If y+b=m_1(x+a)& y+b=m_2(x+a) are two tangents to the parabola y^2=4ax then |m_1m_2| is equal to:

If y+a=m_(1)(x+3a),y+a=m_(2)(x+3a) are two tangents to the parabola y^(2)=4ax , then

5.If y+3=m_(1)(x+2) and y+3=m_(2)(x+2) are two tangents to the parabola y^(2)=8x, then

If y+3=m_1(x+2) and y+3=m_2(x+2) are two tangents to the parabola y_2=8x , then (a)m_1+m_2=0 (b) m_1+m_2=-1 (c)m_1+m_2=1 (d) none of these

If y=m_(1)x+c and y=m_(2)x+c are two tangents to the parabola y^(2)+4a(x+a)=0 then (a)m_(1)+m_(2)=0(b)1+m_(1)+m_(2)=0 (c) m_(1)m_(2)-1=0 (d) 1+m_(1)m_(2)=0