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  • If is the normal at a point in the parabola <img alt="\mathrm{y^2=5 x}" src="/latex-image/?%5Cmathrm%7By%5E2%3D5%20x%

If \mathrm{y=m+c}is the normal at a point in the parabola \mathrm{y^2=5 x} whose focal distance is 8 unit, then \mathrm{\frac{|c|}{\sqrt{3}}} is equal to
 

Option: 1

8


Option: 2

2


Option: 3

16


Option: 4

10


Answers (1)

best_answer

\mathrm{If \: y=m x+c \text{is the normal to the parabola} \: y^2=8 x, then c=a t^3+2 a t\: and \: m=-t}

\mathrm{Here, a=2}

Given, focal distance is 8 units.

\mathrm{ \therefore\left(a t^2-a\right)^2+4 a^2 t^2=64 }

\mathrm{ \Rightarrow a^2\left[\left(t^2-1\right)^2+4 t^2\right]=64 \Rightarrow a^2\left(t^2+1\right)^2=64 }

\mathrm{ \Rightarrow\left(a\left(t^2+1\right)\right)=8 \Rightarrow t^2+1=4 \Rightarrow t^2=3 \Rightarrow t= \pm \sqrt{3} }

\mathrm{ Now, c=a t^3+2 a t}

\mathrm{ = 2(\sqrt{3})^3+2(2) \sqrt{3}=6 \sqrt{3}+4 \sqrt{3}=10 \sqrt{3} }

\mathrm{ \Rightarrow \quad c = \pm 10 \sqrt{3} \quad \therefore \quad|c|=10 \sqrt{3} }

Hence option 4 is correct.








 

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