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.
Answer
383.4k+ views
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.
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:
\[{{c}^{2}}=4ax\]
\[\Rightarrow x=\dfrac{{{c}^{2}}}{4a}\]
So, the point \[P\] is \[\left( \dfrac{{{c}^{2}}}{4a},c \right)\].
Substituting \[y=c\] in \[y=-4bx\], we get:
\[{{c}^{2}}=-4ax\]
\[\Rightarrow x=\dfrac{-{{c}^{2}}}{4b}\]
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
\[4a=4b\]
\[\Rightarrow a=b\]
\[\Rightarrow b-a=0\]
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 \[\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2},\dfrac{{{y}_{1}}-{{y}_{2}}}{2} \right)\].
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.

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:
\[{{c}^{2}}=4ax\]
\[\Rightarrow x=\dfrac{{{c}^{2}}}{4a}\]
So, the point \[P\] is \[\left( \dfrac{{{c}^{2}}}{4a},c \right)\].
Substituting \[y=c\] in \[y=-4bx\], we get:
\[{{c}^{2}}=-4ax\]
\[\Rightarrow x=\dfrac{-{{c}^{2}}}{4b}\]
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
\[4a=4b\]
\[\Rightarrow a=b\]
\[\Rightarrow b-a=0\]
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 \[\left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2},\dfrac{{{y}_{1}}-{{y}_{2}}}{2} \right)\].
Recently Updated Pages
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
What is 1 divided by 0 class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many crores make 10 million class 7 maths CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

State the laws of reflection of light

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
