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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.

Samar Imam Zaidi
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  • $$5,\left(-\left({{2,y}\over{\sqrt{5}}}+{{x}\over{\sqrt{5}}}+{{3, \sqrt{5}+2}\over{5}}\right)^2+\left(y-{{4,\sqrt{5}-25}\over{25}} \right)^2+\left(x-{{2,\sqrt{5}-25}\over{25}}\right)^2\right)$$ – Jan-Magnus Økland Dec 12 '18 at 06:11

3 Answers3

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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 $(1),(2)$

lab bhattacharjee
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  • I did according to your answer Equation of tangents $tY=X+at^2$ and equation of normal passing through $(a,0)$ is $Y=-tX+at$ but i am getting complicate equation but according to me answer must come as line because locus should be tangent at vertex. – mathophile Feb 05 '23 at 11:34
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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.

sonu
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  • This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation. – amd Dec 12 '18 at 07:54
  • I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that. – sonu Dec 12 '18 at 07:58
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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.

amd
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