Image of the conic $\frac{x^2}{25} + \frac{y^2}{16}=1$ in the line $x+y=10$ is :
So for this , since it's an ellipse , I thought of taking the images of the foci and since the e of the ellipse will remain the same , I will form the equation through it.
But , when I saw the solution , all it did was replace $x \to y -10$ and $y \to x-10$ , to form the image. I couldn't understand what exactly did the solution do ?? It gives the exact same answer with far fewer steps. Please help
Also it seems a bit weird as if it had been a line like $x+2y=10$, putting $x \to 10-2y ,$ and $2y \to 10-x $ would have changed the eccentricity which should not have happened
Also is this valid for any general conic ?? Because if it had not been an ellipse and just some random conic , it would have become really hard through my method.
Can I even extend this further , to a general function ??
You just need to find the reflection $(x',y')$ of a general point $(x,y)$ along the given line. It comes out to be $(x',y')=(10-y,10-x)$ or $(x,y)=(10-y',10-x')$. Now just substitute for $x,y$ in the equation of the curve.
For your second question, $(10-2y,10-x)$ is not the reflection of $(x,y)$ along $x+2y=10$, so you will not obtain the desired curve.
Yes, you can extend this procedure for a general curve $y=f(x)$ and a general transformation, provided you are able to find explicitly $x=h(x',y'),y=g(x',y')$. The transformed curve would then be given by$$g(x',y')-(f\circ h)(x',y')=0$$