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QUESTION:

The points of intersection of the curves whose parametric equation are x=t^2+1,y=2t and x=2s and y=2/s is given by:

Options:

a)(1,-3)

b)(2,2)

c)(-2,4)

d)(1,2)

MyApproach:

I have tried to find out the value of t and s by equating the values of x and y. by equating the values of x,i have got a relation 2s^3=1+s^2 and by equating the values of y,i have got t=1/s.I cannot further proceed from here.

Conclusion:

A solution to this problem would be very useful.

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    Hint: The equation $2s^3 = 1 + s^2$ has a rational root (an easy one). – quasi Jan 28 '18 at 07:50
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    Since it is a objective question i guess you have around 30 sec to 1 min to solve this. The point of intersection lies on both curve and the second curve leds to xy=4 and only one option satisfies it – NewGuy Jan 28 '18 at 07:59

1 Answers1

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Eliminating the parameter in both cases seems to be a good approach. In the first set of equations, if $y = 2t$, then $t = \frac{1}{2}y$. Substituting that into the corresponding $x$ equation gives $x = \frac{1}{4}y^2 + 1$. We do the same thing with the second set of equations, and this gives us $y = 4/x$. Then you can either test the points you're given to find the intersection, or you could plug $y=4/x$ into $x = \frac{1}{4}y^2 + 1$ and solve for $x$.

ArbitraryConstant
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  • If you just want to test the points, on an exam say, it's easier to do it directly. We reject a) and d) because $t^2+1=1$ gives $t=0,$ and we know $t=1/s.$ We reject c) because $t^2+1 \ne -2.$ Done. The point of the problem though, comes after what you've done. When you plug in the value for $y,$ you get a cubic in $x$. – saulspatz Jan 28 '18 at 08:00