First, after expressing $(x,y)$ in polar coordinates, you get the equation $(r\cos t+1)^2+(r\sin t)^2=1$, which after a few computations turns out to be $r(r+2\cos t)=0$. So, $r=0$(which is the case when $(x,y)=(0,0)$, we are not interested in this) or $(r+2\cos t)=0$, i.e. $r=-2\cos(t)$. So, the parametrization that we get is $$(x,y)=(-2\cos^2(t),-2\cos t\sin t)$$
(so the parametrization that you got was missing a "-" sign.)
Now, we are just left with finding the bounds for $t$ (i.e. find the interval). We want $A=(-2,0)$ and $B=(-1,1)$ to be in our parametrization. Note that $A$ occurs when $t=0,\pi,2\pi,\cdots$ and $B$ occurs when $t=\pi/4,3\pi/4,\cdots$. Now, you want your parametrization not containing $(0,0)$ (which occurs when $t=\pi/2,3\pi/2,\cdots$;
so you just need to choose a good interval not containing these $t$-values.
For instance $t\in[0,\pi/4]$ works.
Why is the graph below impossible?
Let $\displaystyle m_1=\min_{p\in[p_1,p_2]}f(p)$ and $\displaystyle m_2=\max_{p\in[p_1,p_2]}f(p)$. WLOG, let the two outer intersection points in the graph be $p_1$ and $p_2$ from left to right, i.e., $f(p_1)=f(p_2)=m_2$. Let the inner two points be $A$ and $B$ from left to right, i.e., $f(A)=f(B)=m_1$. We have $m_1<m_2$; otherwise, the line intersects the level curve $f(x,y)=m_1$ at those four points.
Let $E$ be the midpoint between $A$ and $B$.
- $f(E) = m_1$. The level curve that passes through $E$ also passes $A$ and $B$.
- $f(E) > m_1$. The level curve $f(x,y)=\min(f(E),m_2)$ passes a point between $p_1$ and $A$ that is not $A$, a point between $A$ and $E$, a point between $B$ and $p_2$.
However, any strictly convex curve intersects with any line at no more than $2$ points. Hence, the graph above is impossible.
Please see a complete proof below, anyway, if you read the graph differently.
A complete proof
Given point $A\in\mathbb R^2$, let its coordinate be $(A_x, A_y)$ and $f(A)=f(A_x,A_y)$ by slight abuse of $f$. Let $\mathcal C_A$ be the level curve of $f$ that passes $A$, i.e. the curve defined by the equation $f(x,y)=f(A)$.
Pick an abitrary point $P\in \mathcal L$. Since $P\in \mathcal L\cap \mathcal C_P$, by assumption, $|\mathcal L\cap \mathcal C_P|=2$ or $\mathcal L$ is tangent to $\mathcal C_P$. If it is the latter case, we are done.
Otherwise, $|\mathcal L\cap \mathcal C_P|=2$. Let $Q\in \mathcal L\cap \mathcal C_P$, $Q\not=P$. WLOG, suppose $P_x<Q_x$; otherwise, just switch $P$ and $Q$.
$f$ on line segment $\overline{PQ}$ cannot be constant; otherwise, $\mathcal L\cap\mathcal C_P$ would contain infinitely many points. There are two cases.
The minimum value of $f$ on $\overline{PQ}$ is smaller than $f(P)$.
Suppose the minimum value is reached at $S$ for some $S\in\overline{PQ}$. Then $f(S)\lt f(P)$, $S\not=P$ and $S\not=Q$.
Claim. Let $B\in \mathcal L$, $B\not=S$. Then $f(S)\lt f(B)$.
Proof. Let $g(x):x\to \mathbb R$, $g(x)=f(\text{the point on $\mathcal L$ with first coordinate }x)$. $g(P_x)=f(P)=f(Q)=g(Q_x)$.
Suppose $f(S)\ge f(B)$ for the sake of contradiction. There are several cases of $B_x$.
$B_x<P_x$.
$\quad\quad\begin{matrix}\mathcal L:\ \\\ \end{matrix} \overline{\quad\quad B\quad\quad P\quad\quad\phantom{B}\quad\phantom{D}\quad S\quad \quad\quad Q\quad\quad\quad}$
Since $g$ is continuous, $g(x)=\frac{f(S)+f(P)}2$ as an equation in $x$ has one root in interval $(B_x, P_x)$, a root in interval $(P_x, S_x)$ and a root in interval $(S_x, Q_x)$. That means $\mathcal L$ intersects the level curve $f(x,y)=\frac{f(S)+f(P)}2$ at three points, which cannot be true.
$B_x = P_x$, which is not possible since $f(S)<f(P)$.
$P_x<B_x<S_x$. Since $g(S_x)$ is the minimum value of $g$ on $[P_x, Q_x]$, $f(S)\le f(B)$. Hence $f(S)=f(B)$.
Let $D$ be the midpoint of $B$ and $S$.
$\quad\quad\begin{matrix}\mathcal L:\ \\\ \end{matrix} \overline{\quad\quad\phantom{B}\quad\quad P\quad\quad B\quad D\quad S\quad\quad\quad Q\quad\quad\quad}$
- If $f(S)=f(D)$, then $\mathcal L$ intersects the level curve $\mathcal C_S$ at $B, D, S$, which cannot be true.
- Otherwise, $f(S) < f(D)$ since $g(S_x)$ is the minimum value of $g$ on $[P_x, Q_x]$. Then $g(x)=\frac{f(S)+\min(f(P), f(D))}2$ as an equation in $x$ has one root in interval $(P_x, B_x)$, one root in interval $(B_x, D_x)$ and one root in interval $(D_x,S_x)$. That cannot be true.
The remaining cases are symmetric to the cases above. $\quad\checkmark$
The claims says $\mathcal L$ intersects $\mathcal C_S$ only at $S$. That means $\mathcal L$ is tangent to $\mathcal C_S$.
The minimum value of $f$ on $\overline{PQ}$ is bigger than $f(P)$.
This case is symmetric to the case above.
Proof is completed.
Best Answer
There are several ways to go about it. Here is a multivariable calculus approach.
Level curves of $f(x,y) = x^4 + y^2$ will be perpendicular to the direction of the gradient of $f$, $$\nabla f = 4x^3\hat{\mathbf{x}} + 2y\hat{\mathbf{y}}.$$ Furthermore, level curves of $g = g(x,y)$ will be perpendicular to level curves of $f$ if $\nabla f$ and $\nabla g$ are everywhere orthogonal. $\nabla f$ and $\nabla g$ are orthogonal if their dot product is zero: \begin{align} 0 &= \nabla f\cdot\nabla g \\ &= 4x^3\frac{\partial g}{\partial x} + 2y\frac{\partial g}{\partial y} \end{align} This is a partial differential equation for the unknown function $g$. This PDE is satisfied when \begin{align} \frac{\partial g}{\partial x} &= \frac{1}{4x^3} \\ \frac{\partial g}{\partial y} &= -\frac{1}{2y} \end{align} After integrating each, it is clear that, up to an arbitrary constant $g$ must be \begin{align} g(x,y) = -\frac{1}{8x^2} - \frac{1}{2}\ln(y) \end{align} All level curves of $g$ will be orthogonal to level curves of $f$; now simply choose the one where $g(1,1) = g(x,y)$.