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Locus of the midpoint of any focal chord of \mathrm{y}^2=4 \mathrm{ax} is
 

Option: 1

\mathrm{y}^2=\mathrm{a}(\mathrm{x}-2 \mathrm{a})
 


Option: 2

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-2 \mathrm{a})

 


Option: 3

\mathrm{y}^2=2 \mathrm{a}(\mathrm{x}-\mathrm{a})
 


Option: 4

\text{None of these}


Answers (1)

best_answer

Let the midpoint be \mathrm{P}(\mathrm{h}, \mathrm{k}). Equation of this chord is \mathrm{\mathrm{T}=\mathrm{S}_1}. i.e., \mathrm{\mathrm{yk}-2 \mathrm{a}(\mathrm{x}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. It must pass through (\mathrm{a}, 0)

\mathrm{\Rightarrow 2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^2-4 \mathrm{ah}}. Thus required locus is \mathrm{\mathrm{y}^2=2 \mathrm{ax}-2 \mathrm{a}^2.}

Hence option 2 is correct.

 

Posted by

SANGALDEEP SINGH

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