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If the normals any point to the parabola `x^(2)=4y` cuts the line y = 2 in points whose abscissar are in A.P., them the slopes of the tangents at the 3 conormal points are in |
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Answer» `y^2=4ax----------` `y=mx-2am-am^3` `x^2=4ay` `x=my-2am-am^3` `a=1` `x=my-2m-m^3` `P(h,k)` `h=mk-2m-m^3` `m^3+2m-mk+h=0` `m^3+(2-k)m+h=0` Cubic equation`m_1,m_2,m_3` `m_1+m_2+m_3=0` `m_1^3+m_2^3+m_3^3=3m_1m_2m_3` `m_1,m_2,m_3->slope of normal` `-1/m_1,-1/m_2,-1/m_3`->slope of tangent at foot of normal `x=my-2m-m^3//y=2` `x=2m-2m-m^3` `x_1=-m_1^3,x_2=-m_2^3,x_3=-m_3^3` `x_1+x_3=2x_2` `-m_1^3-m_3^3=-2m_2^3` `m_1^3+m_3^3=2m_2^3` `2m_2^3+m_2^3=3m_1m_2m_3` `m_2^2=m_1m_3` `1/m_2^2=1/m_1xx1/m_3` `(-1/m_2)^2=-1/m_1xx1/m_3` `b^2=ac` `-1/m_1,-1/m_2,-1/m_3`are in GP option `(b)` |
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