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

Last updated date: 15th Jul 2024
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Hint: First convert the given line equation to the standard form and compare to find the slope of the line. Substitute these values into the standard line equation to prove the required result.

We have to prove the straight line $lx + my + n = 0$ is tangent to parabola $y^2 = 4a(x - b)$.

If line $y = Mx + c$ touches parabola $y^2 = 4a(x - b)$ then

$c = \dfrac{a}{M}{\text{ }}....................{\text{(1)}}$

For the given line $lx + my + n = 0$

$l(x + b) + my + n = 0$

$y = \dfrac{{ - l(x + b) - n}}{m}$

$y = \dfrac{{ - lx}}{m} + \dfrac{{ - lb - n}}{m}....................{\text{(2)}}$

Compare equation (2) with the equation $y = Mx + c$

$M = \dfrac{{ - l}}{m},c = \dfrac{{ - lb - n}}{m}$

Put these value in the equation (1) the equation become

$\dfrac{{ - lb - n}}{m} = \dfrac{a}{{\dfrac{{ - l}}{m}}}$

$\dfrac{{ - lb - n}}{m} = \dfrac{{am}}{{ - l}}$

$l{b^2} + nl = a{m^2}$

Hence Proved.

Note: If $l{b^2} + nl = a{m^2}$ then the line lx + my + n = 0 will touches the parabola

${y^2} = 4a(x - b).$