Analytic geometry

Geometry Level 1

The curve described parametrically by { x = t 2 + t y = t 2 − t \begin{cases} x=t^{2}+t \\ y=t^{2}-t \end{cases} represents which of the following shapes?

an ellipse a hyperbola a parabola a pair of straight lines

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Aman Sharma by Aman Sharma , 25, India

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1 solution

It is given that:

{ x = t 2 + t y = t 2 − t ⇒ { x + y = 2 t 2 x − y = 2 t ⇒ ( x − y ) 2 = 4 t 2 = 2 ( x + y ) \begin{cases} x=t^2+t \\ y=t^2-t \end {cases} \quad \Rightarrow \begin{cases} x+y = 2t^2 \\ x-y=2t \end {cases} \quad \Rightarrow (x-y)^2=4t^2=2(x+y)

⇒ x 2 − 2 x y + y 2 − 2 x − 2 y = 0 \Rightarrow x^2 - 2xy +y^2 - 2x - 2y = 0

The equation is a conic of form a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 , with a = 1 a = 1 , b = 1 b=1 , c = 0 c=0 , f = − 1 f=-1 , g = − 1 g=-1 and h = − 1 h=-1 .

We note that h 2 − a b = ( − 1 ) 2 − ( 1 ) ( 1 ) = 0 h^2-ab = (-1)^2 - (1)(1) = 0 , which means that the curve is a p a r a b o l a \boxed {parabola} .

For notes on parabola see Conics: Parabola--General

24 Helpful 5 Interesting 7 Brilliant 3 Confused

I did everything correctly, and then made a mistake in the end regarding identifying whether the form was of a parabola or hyperbola ._.

Prasun Biswas - 6 years, 5 months ago

This solution is great!!! .I did mistake that I left the 2 in 2h.

But, next time I will take care.

Wonder full solution.

Soumo Mukherjee - 6 years, 5 months ago

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