I'm trying to solve the following problem:
A pair of tangents are drawn to $x^2+4y^2+4x+3=0$ from points on the parabola $y^2=4x$. The corresponding chords of contact always touch a fixed conic section which is?
I have approached this question as follows
First, I take the parametric point on the parabola, $(t^2,2t)$.
I write the equation for the chord of contact by $T=0$: $$xt^2+8yt+2(x+t^2)+3=0$$ Then, I take the equation of the conic section we need to find as $ax^2+by^2+2gx+2fy+c=0$
I take a point $(h,k)$ on this conic. From this point, the tangent would be, from $T=0$,$$akx+bhy+g(h+x)+f(y+k)+c=0$$
Comparing coefficients for both of these, I get three equations: $$gh+fk+c=2t^2+3\qquad\ [1]\\bk+f=8t \qquad\qquad\qquad\quad[2]\\ah+g=t^2+2\qquad\qquad\quad\ [3]$$ Solving $[2]$ and $[3]$, and replacing $h\rightarrow x$ and $k\rightarrow y$, I get the equation for this conic section:$${\left(\frac{by+f}{8}\right)}^2=ax+(g-2)$$ Which is the equation of a parabola.
However, the given answer is a hyperbola.
Is my method incorrect? What can I do to correct it?
You start out fine; the polar of $(t^2,2t)$ is $x t^2+4y2t+2(x+t^2)+3=0.$ Then the trick to get the envelope is to take the partial derivative wrt $t,$ it is $2tx+8y+4t=0.$ Eliminating $t$ from the two equations gives $2x^2-16y^2+7x+6=0$ or ${\large \frac{(\frac{x+\frac74}{\frac14})^2-(\frac{y}{\frac1{8\sqrt2}})^2-1}8=0}$ which is a hyperbola.
As to what you're doing: you don't use the equation $ah^2+bk^2+2gh+2fk+c=0.$ So the polar is not necessarily the tangent. Adding back in that condition, and considering a needed scalar multiplier, your approach becomes unwieldy.