Locus of a point.

Question

Consider the locus of a moving point P = $(x, y)$ in the plane which satisfies the law $2x^2 = r^2 + r^4$, where $r^2 = x^2 + y^2$. Then only one of the following statements is true. Which one is it?

(a) For every positive real number d, there is a point $(x, y)$ on the locus such that $r = d$.

(b) For every value $d$, $0 < d < 1$, there are exactly four points on the locus, each of which is at a distance $d$ from the origin.

(c) The point P always lies in the first quadrant.

(d) The locus of P is an ellipse.

The answer is option b.

I could eliminate options c and d because it's not an ellipse and $(x,y)$ can be in any quadrant. However, I'm struggling to disprove option 'a'. How do I do that?

Answer 1

Suppose $d=2$, then $r^2=x^2+y^2=4$. In which case the given equation $2x^2=r^2(r^2+1)$ gives, $x^2=10$, but then $y^2=r^2-x^2$ implies $y^2=-6$ which is impossible.

In fact it can be shown that if $d>1$, then you cannot have any point $(x,y)$ satisfying the given equation.

Answer 2

Let $r=d>1$. Then $r^2>1$ and so $r^4>r^2$, hence $2x^2=r^2+r^4>r^2+r^2=2r^2$. Therefore $x^2+y^2\geq x^2>r^2$ which means no such $(x,y)$ is on the curve.

Answer 3

(b) Given $0<d<1$, we find exactly two solutions for $x$ in $2x^2=d^2+d^4$ because the right hand side is positive. Likewise, we find two solutions for $y$ in $2y^2=2(d^2-x^2)=d^2-d^4$, again because the right hand side is positive. By construction, we have $x^2+y^2=d^2$ in all four combined cases.

(a) Note that we found $2y^2=d^2-d^4$ in the proof for (b). For $d>1$ the right hand side is negative.

Answer 4

With $x=r\cos \theta$ the equation $2x^2 = r^2 + r^4$ becomes
$$2r^2\cos^2\theta=r^2+r^4$$ or, for $r\neq 0,$ $$2\cos^2\theta=1+r^2.$$Here $LHS\leq2$ and consequently $r^2\leq1.$
Thus if $d>1,$ no value of $r$ satisfies.

Answer 5

I would like here to underline that the given polar equation can be rewritten in a simpler form giving rise to a lemniscate of Bernoulli (see figure below and https://www.mathcurve.com/courbes2d.gb/lemniscate/lemniscate.shtml).

Indeed, replacing $x$ by $r \cos \theta$ gives

$$2r^2\cos^2\theta=r^2+r^4 \iff \ \begin{cases}r^2&=&0\\2\cos^2\theta&=&1+r^2\end{cases}$$

i.e., the polar equation of the union of two curves ; the first one, $r^2=0$, represents in fact a single point, the origin, whereas the other can be given the form

$$r^2=2 \cos^2 \theta - 1$$

(same transformation as done by @user376343) with a further simplification into

$$r^2=\cos(2\theta)\tag{1}$$

We recognize in (1) the polar equation of a lemniscate of Bernoulli.

Such a curve is bounded, contradicting option (a).