A) \[a{{l}^{3}}-2al{{m}^{2}}={{m}^{2}}\]
B) \[a{{l}^{2}}+2al{{m}^{3}}={{m}^{2}}\]
C) \[a{{l}^{3}}+2al{{m}^{2}}={{m}^{3}}\]
D) \[a{{l}^{3}}+2al{{m}^{2}}={{m}^{2}}\]
Correct Answer: D
Solution :
Let \[P\,({{x}_{1}},\,{{y}_{1}})\] be a point on the curve \[{{y}^{2}}=4ax\] ?(i) On differentiating \[{{y}^{2}}=4ax\]w.r.t. x, we get \[2y\frac{dy}{dx}=4a\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=\frac{2a}{{{y}_{1}}}\] Thus, the equation of normal at \[({{x}_{1}},\,{{y}_{1}})\]is \[y-{{y}_{1}}=-\frac{{{y}_{1}}}{2a}(x-{{x}_{1}})\] \[\Rightarrow \] \[{{y}_{1}}x+2ay={{y}_{1}}({{x}_{1}}+2a)\] ?(ii) But \[lx+my=1\] ?(iii) is also a normal. Therefore, coefficients of Eqs. (ii) and (iii), must be proportional. \[\therefore \] \[\frac{{{y}_{1}}}{l}=\frac{2a}{m}=\frac{{{y}_{1}}({{x}_{1}}+2a)}{1}\] \[\Rightarrow \] \[{{y}_{1}}=\frac{2al}{m}\] and \[{{x}_{1}}=\frac{1}{l}-2a\] Putting these values of \[{{x}_{1}}\] and \[{{y}_{1}}\] in Eq. (i), we get \[{{\left( \frac{2al}{m} \right)}^{2}}=4a\left( \frac{1}{l}-2a \right)\] \[\Rightarrow \] \[\frac{4{{a}^{2}}{{l}^{2}}}{{{m}^{2}}}=\frac{4a-8{{a}^{2}}l}{l}\] \[\Rightarrow \] \[a{{l}^{3}}={{m}^{2}}-2a{{m}^{2}}l\] \[\Rightarrow \] \[a{{l}^{3}}+2al{{m}^{2}}={{m}^{2}}\] Alternate We know that the equation of normal to the curve \[{{y}^{2}}=4ax\] in slope form is \[y=Mx-2aM-a{{M}^{3}}\] ?(i) Also, given equation of normal to\[~{{y}^{2}}=4ax,\]is \[y=-\frac{l}{m}x+\frac{1}{m}\] ?(ii) On comparing Eqs. (i) and (ii), we get \[M=-\frac{l}{m}\] ?(iii) and \[-2aM-a{{M}^{3}}=\frac{1}{m}\] ?(iv) From Eqs. (iii) and (iv), we get \[-2a\left( -\frac{l}{m} \right)-a{{\left( -\frac{l}{m} \right)}^{3}}=\frac{1}{m}\] \[\Rightarrow \] \[2al{{m}^{2}}+a{{l}^{3}}={{m}^{2}}\]
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