<s>[INST] From a point on a given circle, tangents are drawn to the ellipse. Need to find locus of chord of contact. From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$
I recognize that the locus of a chord whose midpoint is at $(h,k)$ is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$
I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.
Here I am getting the equation in terms of $x,y,h,k$, but to find the locus I need the equation entirely in the form of $h,k$, right? So how do I eliminate $x,y$ from the equation of the locus of the midpoint?
 [/INST] This may not be the solution you are looking for:
Let $\mathcal C=\{x^2+y^2=1\}$ be the unit circle. Let $O' = (\alpha, \beta)$ be any point outside of this circle. Let $O'P'$ and $O'Q'$ be two tangent line to $\mathcal C$. One can check that the midpoint $m' = (x, y)$ of $P'Q'$ is given by (why?)
$$(x, y) = m' = \frac{1}{\alpha^2+ \beta^2} (\alpha, \beta).$$
Now assume that $O = (\alpha, \beta)$ is on the ellipse $\{ (ax)^2 + (by)^2 = d^2\}$. Thus $(a\alpha)^2 + (b\beta)^2 = d^2$. Then 
$$ d^2(x^2 + y^2)^2 = \frac{d^2}{(\alpha^2 + \beta^2)^2}$$
and 
$$(ax)^2 +(by)^2 = \frac{(a\alpha)^2 + (by)^2}{(\alpha^2 + \beta^2)^2}=\frac{d^2}{(\alpha^2 + \beta^2)^2}$$
Thus the locus of the midpoint $m'$ is given by
$$ \tag{1} (ax)^2 + (by)^2 = d^2 (x^2+ y^2)^2. $$
The above is related to your question in the following way: Consider the transformation: $$(x, y) \mapsto (x/a, y/b).$$ 
Under this transformation, the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ is sent to the unit circle $\mathcal C$, while the circle $x^2 + y^2 = d^2$ is sent to the ellipse $(ax)^2 + (by)^2 = d^2$. The crucial observation is that tangent lines $OP, OQ$ are also sent to tangent lines $O'P', O'Q'$, and the midpoint $m$ of $PQ$ are sent to the midpoint $m'$ of $P'Q'$ (see here). Thus if you take the inverse transformation 
$$ (x, y) \mapsto (ax, by)$$
Then the locus of $m'$ will be sent to the locus of $m$. This implies your equation: if you change $x$, $y$ to $x/a$, $y/b$ respective in (1), you get 
$$x^2 + y^2 = d^2 \left(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\right)^2.$$
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