conic sections
Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=\frac{8}{\sqrt{5}}(x+2y+3)$.
My approach
I am trying to convert above equation in parabolic form
$\frac{(ax+by+c)^2}{a^2+b^2}=(x-\alpha)^2+(y-\beta)^2$
where $ ax+by+c=0$ is the equation of directrix and ($\alpha,\beta$) is the focus of the parabola but getting complicated.
I think it is a circle whose radius is $a$ and center is the origin.
As is shown in the graph, suppose C is a point on the ellipse, and F1,F2 are two focus, CM is the angle bisector of $\angle F_1CF_2$, and F2P is perpendicular to the tangent line, E lies at both line F1C and F2P.
According to the reflexive property, it is easy to know that CF2 = CE, and hence $\triangle CF_2P$ is congruent with $\triangle CEP$, so $P$ is the mid point of $F_2E$ and $O$ is the midpoint of $F_1F_2$, so $OP$ is always half the length of $F_1E$, and we know that $ CF_1 + CE = CF_1 + CF_2 = 2a$, so $OP = a$.
To find the vertex of that parabola, you can make use of a nice general result:
given three tangents of a parabola, if $AB$ is that part of a tangent which is comprised between the other two tangents (see diagram), then the projection of $AB$ onto a line perpendicular to the axis has a fixed length, independent of the position of $AB$.
In our particular case, if $V$ (vertex) and $D$ are the tangency points of the "outer" tangents, $C$ is their intersection point, and $B'$, $D'$ are the projections of $B$, $D$ on tangent $VC$ (which is perpendicular to the axis), we obtain from the above result:
$$AB'=CD'=VC.$$
As points $A$, $B$, $C$, $B'$ can be readily found from the data, one can also easily find vertex $V$ and tangency point $D$.
Best Answer
Hint :
Let $Y=2x-y+1\ \ \ \ (1)$ and $4aX=\dfrac{8(x+2y+3)}{\sqrt5}$ with $x+2y+3=X\ \ \ \ (2), a=?$
whose focus is $(a,0)$
Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$
Find the equation of the tangent at $P$
Find the equation of the normal for the tangent passing through $(a,0)$
Find the intersection of the normal with the tangent.
Eliminate $t$
Replace the values of $X,Y$ with $x,y$ using $(1),(2)$