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  • The locus of the mid-point of the line segment joining the focus to a moving point on the parabola  <img alt="\mathrm {y^2=4 a x }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola  \mathrm {y^2=4 a x }  is another parabola with directrix
 

Option: 1

\mathrm { x=-a }


Option: 2

\mathrm {x=-a / 2}


Option: 3

\mathrm {x=0}


Option: 4

\mathrm {x=a / 2}


Answers (1)

best_answer

The focus of parabola \mathrm {y^2=4 a x \: \: is\: \: S(a, 0).} 


Let   \mathrm {P\left(a t^2, 2 a t\right)}  be any point on the parabola. The midpoint of   \mathrm {S P }  is given by


\begin{aligned} & \mathrm {x=\frac{a\left(t^2+1\right)}{2}, y=\frac{2 a t+0}{2}=a t }\\ \Rightarrow & 2\mathrm { x=a\left[\frac{y^2}{a^2}+1\right]=\frac{y^2}{a}+a }\\ \Rightarrow & \mathrm {y^2=2 a x-a^2} \\ \Rightarrow & \mathrm {y^2=2 a\left(x-\frac{a}{2}\right)} \\ \Rightarrow & \mathrm {y^2=4 A X\{4 A=2 a, A=a / 2\}} \end{aligned}

which is parabola with directrix  \mathrm {X=-A,} 


\Rightarrow \begin{aligned} & \mathrm {x-a / 2=-a / 2 }\\ & \mathrm x=0 \end{aligned}

Posted by

sudhir.kumar

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