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  • Consider the two parabolas <img alt="\mathrm{y^2=4 a x \, \, and\, \, y^2=4 c(x-b)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%5E2%3D4%20a%20x%20%5C%2C%20%5C%2C%20and%5C%2C

Consider the two parabolas \mathrm{y^2=4 a x \, \, and\, \, y^2=4 c(x-b)}. The two parabolas have exactly one commom normal if:

Option: 1


a=c 


Option: 2

 a=2 c

 


Option: 3

2 a=c


Option: 4

 a = -c


Answers (1)

best_answer

\mathrm{y^2=4 a x..............(1) }
Equation of the tangent to (1) at the point \mathrm{\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right) : }
\mathrm{\begin{aligned} & y=m x+\frac{a}{m}............(2) \\ & y^2=4 c(x-b).....................(3) \end{aligned} }
Equation of tangent having same slope as that of (2), to (3) at the point \mathrm{\left(\frac{\mathrm{c}}{\mathrm{m}^2}+\mathrm{b}, \frac{2 \mathrm{c}}{\mathrm{m}}\right) }
\mathrm{\mathrm{y}=\mathrm{m}(\mathrm{x}-\mathrm{b})+\frac{\mathrm{c}}{\mathrm{m}}...............(4) }
Since the line joining the points of tangency shall be common normal,
\mathrm{\begin{aligned} & \frac{\frac{2 a}{m}-\frac{2 c}{m}}{\frac{a}{m^2}-\left(\frac{c}{m^2}+b\right)}=-\frac{1}{m}..................(5) \\ \Rightarrow \quad & \frac{2(a-c) m}{(a-c)-m^2 b}=-\frac{1}{m} \\ \Rightarrow \quad & 2(a-c) m^2=m^2 b-(a-c) \\ \Rightarrow \quad & m^2=\frac{a-c}{b-2(a-c)}=\frac{1}{\frac{b}{a-c}-2} \end{aligned} }
Hence \mathrm{\mathrm{m} } will have two distinct real values if \mathrm{\frac{\mathrm{b}}{\mathrm{a}-\mathrm{c}}>2 }
Thus for each value of m there will be a corresponding value \mathrm{-\frac{1}{m} } and hence there will be three common normals possible \mathrm{(including \, \, \, \mathrm{y}=0 ). If \mathrm{a}=\mathrm{c}\, \, then \, \, from\, \, (5) -\frac{1}{\mathrm{~m}}=0 } which is the slope of the \mathrm{\mathrm{x}-axis } i.e., exactly one common normal is possible.

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SANGALDEEP SINGH

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