Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=\frac{8}{\sqrt{5}}(x+2y+3)$

Question

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.

Answer 1

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 $(1),(2)$

Answer 2

There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.

Answer 3

The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.

That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.