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I was given my Math C assignment today and the moment I looked at question 1 I knew I had no idea what to do. This is the graph I was given:

I was asked to provide an equation for the curve however I don't understand how you can derive an equation of this because I have never seen anything like it. I thought that the equation would be something such as $$x=\pm |y^2|$$ however this was just a guess after looking at it.

Also there is another question which asks to show that the equation of the chord of $PQ$ is given by $$(t_1+2_2)y-(t_1^2+t_1\times t_2+2_2^2)x+t_1^2\times t_2^2=0$$

And lastly, show that the equation of the tangent to the curve at a point corresponding to $t$, where $t$ doesn't equal $0$, is given by $$2y-3tx+t^3=0$$

Can anyone help me or at least explain what I have to do?

Also for the parametric equation, $$x=t^2 \text{ and }=t^3.$$

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  • $\begingroup$ One question per post, please. $\endgroup$ – Yves Daoust May 18 '15 at 9:49
  • $\begingroup$ For an $x=t^2$ there exists $y= \pm t^3$ , eliminate $t$. i.,e $x^3 = y^2$ , i.e $y= \pm x^{\frac{3}{2}}$ desmos.com/calculator/yznjflvemz $\endgroup$ – Mann May 18 '15 at 9:51
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The parametric equation of the curve is $$x=t^2,\\y=t^3,$$with $t$ taking positive as well as negative values.

You can eliminate $t$ to get an explicit equation $y=f(x)$ by noting that

$$x^3=t^6=y^2.$$

Note that this curve is called a semicubical parabola and is not a conic section.

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  • $\begingroup$ how did you come about x^3 = t^6 = y^2?T Thanks for helping :) $\endgroup$ – Brayden May 18 '15 at 22:15
  • $\begingroup$ Just by seeing that $x$ and $y$ are powers of $t$ and reducing to the same power. $\endgroup$ – Yves Daoust May 19 '15 at 6:36

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