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If \mathrm{f(x+y)=f(x) \cdot f(y)} for all x, y and \mathrm{f(1)=2} and \mathrm{a_r=f(r)}, for \mathrm{r \in N} then the coordinates of a point on the  parabola \mathrm{y^2=8 x} whose focal distance is 4, may be 

Option: 1

\mathrm{\left(a_1, a_2\right)}


Option: 2

\mathrm{\left(a_1,-a_2\right)}

 


Option: 3

\mathrm{\left(a_1,a_1\right)}

 


Option: 4

none of these 


Answers (1)

best_answer

Given parabola is \mathrm{y^{2}=8x}

Here, \mathrm{a=2} let \mathrm{P\left(2 t^2, 4 t\right)} be a point on parabola     ...(i)

Eq. (i), and S be the focus

Given, 

                  \\\mathrm{SP=4}\\\mathrm{\begin{aligned} \\& \therefore \quad a\left(1+t^2\right)=4 \\ \\& 2\left(1+t^2\right)=4 \Rightarrow t= \pm 1 \\ & \end{aligned}}

\therefore                                        P \equiv(2,4) \text { or }(2,-4)

Given, 

\mathrm{f(x+y)=f(x) f(y) \text { for all } x \text { and } y}\: \: \: \: \: \: \: \: \: \: ...(ii)

Given, \mathrm{f(1)=2}

From Eq. \mathrm{\text { (ii), } f(2)=f(1+1)=f(1) \cdot f(1)=2^2=4}

Similarly \mathrm{f(n)=2^{n}}

               \mathrm{\begin{aligned} & a_r=f(r) \\ \\\therefore & \: \: a_1=f(1)=2 \\ \\& a_2=f(2)=4 \end{aligned}}

Hence, \mathrm{P \equiv\left(a_1, a_2\right) \text { or }\left(a_1,-a_2\right)}

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SANGALDEEP SINGH

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