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, $$x=t^2 \text{ and }=t^3.$$

  • $\begingroup$ 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

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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.