This is a second follow-up to [this question](https://math.stackexchange.com/questions/4054110/find-cross-tangents-to-offset-ellipses/4060630#4060630). For the ellipse centered at $(h,k)$ defined implicitly as

$$
\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=0 \tag{1}
$$

Please show the steps for calculating its dual curve and provide the equation for the dual curve. **Do not solve by redefining the function parametrically.** I know how to solve for parametric functions, but need to see the general algorithm for implicit functions.

[Wikipedia](https://en.wikipedia.org/wiki/Dual_curve#Polar_reciprocal) provides a trivial example that does not help me.

**My Attempt**

At point $(p,q)$,

$$
X = \lambda \frac{2(p-h)}{a^2} \implies p = h + \left( \frac{a^2}{2\lambda} \right) X
$$

$$
Y = \lambda \frac{2(q-k)}{b^2} \implies q = k + \left( \frac{b^2}{2\lambda} \right) Y
$$

For incidence to occur, the inner product of the line $[X,Y]$ and point $(p,q)$ must be zero:

$$
\frac{a^2}{2\lambda}X^2 + hX + \frac{b^2}{2\lambda}Y^2 + kY = 0
$$

$$
a^2X^2 + b^2Y^2 + 2\lambda (hX + kY) = 0 \tag{2}
$$

How should $\lambda$ be eliminated at this point?

**Thought 1:**

Can we homogenize the linear terms by $\frac{\left( hX + kY \right)}{-2\lambda}$? If so, we get

$$
a^2X^2 + b^2Y^2 - (hX+kY)^2 = 0
$$

$$
a^2X^2 + b^2Y^2 - (h^2X^2 + 2hkXY + k^2Y^2) = 0
$$

$$
(a^2-h^2)X^2 + (b^2-k^2)Y^2 - 2hkXY = 0
$$

but I'm not sure if this is the correct equation for the dual. Is there a way to check?

**Thought 2:**

Alternatively, can I just say that $\lambda = 1$? In this case, I get:

$$
a^2X^2+b^2Y^2+2(hX+kY)=0
$$

However, now my answer in Thought 1 is different from Thought 2.