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

Prove that the line lx+my+n=0lx + my + n = 0 touches the parabola y2=4a(xb)y^2 = 4a(x - b) if am2=bl2+nla{m^2} = b{l^2} + nl.

Explanation

Solution

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.

Complete step by step answer:

We have to prove the straight line lx+my+n=0lx + my + n = 0 is tangent to parabola y2=4a(xb)y^2 = 4a(x - b).

If line y=Mx+cy = Mx + c touches parabola y2=4a(xb)y^2 = 4a(x - b) then

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

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

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

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

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

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

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

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

lbnm=alm\dfrac{{ - lb - n}}{m} = \dfrac{a}{{\dfrac{{ - l}}{m}}}

lbnm=aml\dfrac{{ - lb - n}}{m} = \dfrac{{am}}{{ - l}}

lb2+nl=am2 l{b^2} + nl = a{m^2}

Hence Proved.

Note: If lb2+nl=am2l{b^2} + nl = a{m^2} then the line lx + my + n = 0 will touches the parabola

y2=4a(xb).{y^2} = 4a(x - b).