A) \[{{y}^{2}}={{({{m}^{1/2}}-{{m}^{-1/2}})}^{2}}ax\]
B) \[{{y}^{2}}={{({{m}^{1/2}}+{{m}^{-1/2}})}^{2}}ax\]
C) \[{{y}^{2}}={{({{m}^{1/2}}+{{m}^{-1/2}})}^{2}}x\]
D) None of these
Correct Answer: B
Solution :
| [b] Consider two points \[P\left( a{{t}^{2}}_{1},2a{{t}_{1}} \right)\] and \[Q\left( a{{t}^{2}}_{2},2a{{t}_{2}} \right)\] on the parabola \[{{y}^{2}}=4ax\] |
| Given: \[\frac{a{{t}^{2}}_{1}}{a{{t}^{2}}_{2}}=\frac{{{m}^{2}}}{1}\]or \[{{t}_{1}}=m{{t}_{2}}\] ? (1) |
| Let \[R(h,k)\] be the point of intersection of tangents at |
| P and Q. |
| Then, \[h=a{{t}_{1}}{{t}_{2}}\] and \[k=a({{t}_{1}}+{{t}_{2}})\] |
| \[\Rightarrow h=am{{t}^{2}}_{2}\] and \[k=a(m{{t}_{2}}+{{t}_{2}})\] |
| [Using (1)] |
| \[\Rightarrow {{t}^{2}}_{2}=\frac{h}{am}\] and \[{{t}_{2}}=\frac{k}{a(m+1)}\] |
| Equating the two values of \[{{t}_{2}},\] we |
| get \[\frac{{{k}^{2}}}{{{a}^{2}}{{(m+1)}^{2}}}=\frac{h}{am}\] |
| \[\Rightarrow {{k}^{2}}=ah\frac{{{(m+1)}^{2}}}{m}\Rightarrow {{k}^{2}}=ah{{\left( \sqrt{m}+\frac{1}{\sqrt{m}} \right)}^{2}}\] |
| \[\therefore \] Required locus is \[{{y}^{2}}=ax({{m}^{\frac{1}{2}}}+{{m}^{\frac{1}{2}}}).\] |
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