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  • The points of intersection of the curves whose parametric equations are  <img alt="x=t^2+1, y=2 t \; \text{and}\; x=2 s, y=\frac{2}{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?x%3

The points of intersection of the curves whose parametric equations are  x=t^2+1, y=2 t \; \text{and}\; x=2 s, y=\frac{2}{s} is given by

Option: 1

(1,-3)
 


Option: 2

(2,2)
 


Option: 3

(-2,4)
 


Option: 4

(1,2)


Answers (1)

best_answer

(b) On eliminating t from x=t^2+1, y=2 t, we get

                         x=\frac{y^2}{4}+1 \Rightarrow y^2=4 x-4\: \: \: \: \: \: \: \: \: \: \: ...(i)

On substituting x=2 s \text { and } y=\frac{2}{s} in Eq. (i), we get 

                      \begin{aligned} \frac{4}{s^2} & =4.2 s-4 \Rightarrow \frac{1}{s^2}=2 s-1 \\ \Rightarrow\: \: \: \: \: \: \: \: 1 & =2 s^3-s^2 \end{aligned}

                      \begin{array}{lr} \Rightarrow & 2 s^3-s^2-1=0 \\ \\\Rightarrow & (s-1)\left(2 s^2+s-1\right)=0 \\ \\\Rightarrow & s=1 \end{array}

                      \therefore \: \: \: \: \: \: \: \: x=2 s=2 \text { and } y=\frac{2}{s}=2

Posted by

HARSH KANKARIA

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