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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{x^2=4 a y}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5

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

Option: 1

\mathrm{y=-a}
 


Option: 2

\mathrm{y=\frac{-a}{2}}
 


Option: 3

\mathrm{y=0}
 


Option: 4

\mathrm{y=\frac{a}{2}}


Answers (1)

best_answer

Focus of the parabola \mathrm{x^2=4 a y \ldots(1)\: is \: (0, a)}

Let \mathrm{P(\alpha, \beta)}be the midpoint of the line segment joining the focus to the variable point \mathrm{(x, y)} on the parabola.

Then \mathrm{\alpha=\frac{x}{2}, \beta=\frac{y+a}{2} \Rightarrow x=2 \alpha, y=2 \beta-a }

\mathrm{\therefore From (1), 4 \alpha^2=4 a(2 \beta-a) }
Required locus is \mathrm{x^2=a(2 y-a)=2 a\left(y-\frac{a}{2}\right) }

Shift the origin to \mathrm{\left(0, \frac{a}{2}\right), x=X, y=Y+\frac{a}{2} }

Then the locus is \mathrm{X^2=2 a Y }

This represents parabola whose directrix is \mathrm{Y=-\frac{a}{2} \: \: or \: \: y=0 }

Hence option 3 is correct.

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Anam Khan

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