\[{\text{Prove that the line }}lx + my + n = 0{\text{ touches the parabola }}{y^2}{\text{ = }}4a(x - b){\text{ if }}a{m^2} = b{l^2} + nl.\]
Answer
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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). \\ $
{\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). \\ $
Last updated date: 24th Sep 2023
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