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Last updated date: 02nd Dec 2023
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MVSAT Dec 2023

Two parabolas have a common axis and concavities in opposite directions. If any line parallel to the common axis meets the parabolas in and , prove that the locus of midpoint of is another parabola, provided that the latera recta of the given parabola are unequal.

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Hint: Equation of line parallel to the common axis is given by \[y=c\].
Let’s consider the two parabolas \[{{y}^{2}}=4ax\] and \[{{y}^{2}}=-4bx\].
From the equation of the parabola, we can see that the axis of the parabolas is common and their concavities are opposite.

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Now, we know the equation of the common axis is \[y=0\].
So, the equation of a line parallel to the common axis is \[y=c\].
Now, this line meets the parabolas \[{y}^{2}=4ax\] and \[{y}^{2}=-4bx\] at \[P\] and \[{{P}^{'}}\] respectively.
Let the midpoint of \[P{{P}^{'}}\] be \[\left( h,k \right)....\left( i \right)\].
Now, substituting \[y=c\] in \[{y}^{2}=4ax\], we get:
\[\Rightarrow x=\dfrac{{{c}^{2}}}{4a}\]
So, the point \[P\] is \[\left( \dfrac{{{c}^{2}}}{4a},c \right)\].
Substituting \[y=c\] in \[y=-4bx\], we get:
\[\Rightarrow x=\dfrac{-{{c}^{2}}}{4b}\]
So, the coordinates of \[{{P}^{'}}\] are \[\left( \dfrac{-{{c}^{2}}}{4b},c \right)\].
Now, know that the coordinates of the midpoint of the line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given as:\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
So, the midpoint of \[P{{P}^{'}}\] is
\[\left( \dfrac{\dfrac{{{c}^{2}}}{4a}-\dfrac{{{c}^{2}}}{4b}}{2},\dfrac{c+c}{2} \right)\]
\[=\left( \dfrac{{{c}^{2}}}{8a}-\dfrac{{{c}^{2}}}{8b},c \right)....\left( ii \right)\]
From \[\left( i \right)\]and \[\left( ii \right)\], we can conclude that
\[h=\dfrac{{{c}^{2}}}{8a}-\dfrac{{{c}^{2}}}{8b}\] and \[k=c\]
Or, \[h=\dfrac{{{k}^{2}}}{8a}-\dfrac{{{k}^{2}}}{8b}\]
Or, \[h=\dfrac{{{k}^{2}}}{8}\left( \dfrac{1}{a}-\dfrac{1}{b} \right)\]
Or, \[h=\dfrac{{{k}^{2}}}{8}\left( \dfrac{b-a}{ab} \right)\]
Or, \[\dfrac{8h\left( ab \right)}{a-b}={{k}^{2}}\]
Or \[{{k}^{2}}=\dfrac{8ab}{b-a}.h\]
Or \[{{k}^{2}}=4\left( \dfrac{2ab}{b-a} \right).h\]
Replacing \[k\] by \[y\]and \[h\] by \[x\], we get:
\[{{y}^{2}}=4\left( \dfrac{2ab}{b-a} \right).x\]
which clearly represents a parabola.
Hence, the locus of the midpoint of \[PP'\] is the parabola \[{{y}^{2}}=4\left( \dfrac{2ab}{b-a} \right).x\].
Now, in case if the latera recta of the parabolas are equal then:
We know, length of latus rectum of parabola \[{{y}^{2}}=4ax\] is \[4a\] and the length of latus rectum of parabola \[{{y}^{2}}=-4bx\] is \[4b\].
If the latera recta are equal, then
\[\Rightarrow a=b\]
\[\Rightarrow b-a=0\]
Now, when we substitute \[b-a=0\] in the equation of locus, we get
 \[{{y}^{2}}=4\left( \dfrac{2ab}{0} \right).x\]
\[\Rightarrow {{y}^{2}}=\infty \] which is an invalid curve.
Hence, for the locus to exist, the latera recta of the two parabolas must be unequal.
 So, the locus of \[P{{P}^{'}}\]is a parabola , given by \[{{y}^{2}}=4\left( \dfrac{2ab}{b-a} \right).x\], which exists only when the latera recta of the given two parabolas are unequal.

Note: Remember that the midpoint of two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]is given as \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]. Students generally make a mistake of writing it as \[\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2},\dfrac{{{y}_{1}}-{{y}_{2}}}{2} \right)\].