A) \[{{l}^{2}}=2mn\]
B) \[l=4{{m}^{2}}{{n}^{2}}\]
C) \[{{m}^{2}}=4ln\]
D) \[{{l}^{2}}=4mn\]
Correct Answer: D
Solution :
Given that \[lx+my+n+0\] .....(i) \[{{x}^{2}}=y\] .....(ii) The point of intersection of the line and parabola are obtained by solving (i) and (ii) simultaneously substituting the values of \[x\] from (i) in (ii), we get \[{{\left( \frac{my+n}{l} \right)}^{2}}=y\] \[\Rightarrow \]\[{{m}^{2}}{{y}^{2}}+{{n}^{2}}+2mny=y{{l}^{2}}\] Þ \[{{m}^{2}}{{y}^{2}}+(2mn-{{l}^{2}})\,y+{{n}^{2}}=0\] ?..(iii) If lines (iii) touches the parabola (ii), then discriminant = 0 Þ \[{{(2mn-{{l}^{2}})}^{2}}=4{{m}^{2}}{{n}^{2}}\] \[\Rightarrow \,4{{m}^{2}}{{n}^{2}}+{{l}^{4}}-4mn{{l}^{2}}=4{{m}^{2}}{{n}^{2}}\]Þ \[{{l}^{2}}=4mn\].
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