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$$P: y^2 = 4x$$ $$L: x+y+4 = 0$$

Find mirror image of P with respect to L and mirror image of L with respect to P.

I know how to find the mirror image of a curve with respect to a line... Mirror image of P with respect to L is $x^2+4y+8x+32=0.$ How to find the mirror image of L with respect to P? Or for that matter mirror image of any curve with respect to a parabola? I'm trying to find some relations for closed form.

EDIT: Someone suggested that this problem was duplicate as another one: find image of parabola with respect to line. Please note that its not. I've seen the solution for that, and understood it.

Sid
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  • I think when it's asked to find the image of a conic with respect to another conic, they generally require us to find the image wrt to the FOCUS of the conic section. – Ishraaq Parvez Mar 01 '21 at 04:11
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    I don't understand what the graph will look like if we try to imagine the mirror image of a line wrt a curve? About which point will you reflect the line in the curve? For finding mirror image of curve wrt line, we can do so by dropping a normal on the line from the curve, and reflect it on the other side. – V.G Mar 01 '21 at 04:15
  • why isnt it possible to find the mirror image with respect to a parabola? theoretically its possible right? draw normal of parabola to line. and the point equidistant on the other side of the normal is the image? – Sid Mar 01 '21 at 04:15
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    I found this. – V.G Mar 01 '21 at 04:20
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    You could also argue "mirror image" of a point P in a smooth curve C gives a curve assuming P is not on C: for every point Q on C reflect P across the tangent at Q. – user10354138 Mar 01 '21 at 04:23
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    @DevashsihKaushik No, the OP already stated he knows how to reflect something across a line. The question is what about reflecting across a parabola. – user10354138 Mar 01 '21 at 04:27

1 Answers1

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A parametric equation for the parabola $y^2 = 4x$ is $P(t) = (t^2, 2t)$

Then the slope of the tangent line to the parabola at the point $P(t)$ is

$$\dfrac{dy}{dx} = \dfrac{2 \, dt}{2t \, dt} = \dfrac 1t$$

Thus the equation of the normal line to the parabola at the point $P(t)$ is

\begin{align} y - 2t &= -t(x - t^2) \\ y-2t &= x -tx + t^3\\ tx + y &= 2t + t^3 \end{align}

This line intersects the line of reflection, $x+y=-4$, at

$$ L(t) = \left(\frac{4 + 2t + t^3}{t-1}, -\frac{6t+t^3}{t-1} \right)$$

Thus, $L'(t)$, the reflection of the point $L(t)$ about the parabola, $P(t)$ is

$$L'(t) = 2P(t) - L(t) = \left(\dfrac{t^3 - 2 t^2 - 2 t - 4}{t - 1}, \dfrac{t^3 + 4 t^2 + 2 t}{t - 1}\right)$$

Steven Alexis Gregory
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