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\[{\text{Prove that the line }}lx + my + n = 0{\text{ touches the parabola }}{y^2}{\text{ = }}4a(x - b){\text{ if }}a{m^2} = b{l^2} + nl.\]

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$ {\text{straight line }}lx + my + n = 0{\text{ is tangent to}} \\
  {\text{parabola }}{y^2} = 4a(x - b) \\
  {\text{if line }}y = Mx + c{\text{ touches parabola }}{y^2} = 4a(x - b) \\
  {\text{then}} \\
  c = \dfrac{a}{M}{\text{ }}....................{\text{(1)}} \\
  {\text{for the given line }}lx + my + n = 0 \\
  l(x + b) + my + n = 0 \\
  y = \dfrac{{ - l(x + b) - n}}{m} \\
  y = \dfrac{{ - lx}}{m} + \dfrac{{ - lb - n}}{m}....................{\text{(2)}} \\
  {\text{compare equation (2) with the equation }}y = Mx + c \\
  M = \dfrac{{ - l}}{m},c = \dfrac{{ - lb - n}}{m} \\
  {\text{put these value in the equation (1) the equation become }} \\
  \dfrac{{ - lb - n}}{m} = \dfrac{a}{{\dfrac{{ - l}}{m}}} \\
  \dfrac{{ - lb - n}}{m} = \dfrac{{am}}{{ - l}} \\
  l{b^2} + nl = a{m^2}.{\text{ Hence Proved}}{\text{.}} \\
  {\text{Note: If }}l{b^2} + nl = a{m^2}{\text{ then the line }}lx + my + n = 0{\text{ will touches the parabola }} \\
  {y^2} = 4a(x - b). \\ $