What Is the Equation of the Curve Reflected Over the Line x-y-2=0?

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SUMMARY

The equation of the curve reflected over the line x - y - 2 = 0 can be derived from the ellipse defined by \(\frac{(x-4)^2}{16} + \frac{(y-3)^2}{9} = 1\). The general point on the ellipse is represented as P(4 + 4cosθ, 3 + 3sinθ). To find the reflected point (h, k), the condition that the line joining P and (h, k) is perpendicular to the line x - y - 2 = 0 must be utilized, leading to a straightforward calculation of h and k in terms of θ.

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Homework Statement


Find equation of the curve on reflection of the ellipse \dfrac{(x-4)^2}{16} + \dfrac{(y-3)^2}{9} = 1 about the line x-y-2=0.

Homework Equations



The Attempt at a Solution


Let the general point be P(4+4cosθ,3+3sinθ). Let the reflected point be (h,k).

\dfrac{4+4cos \theta + h}{2} - \dfrac{3+3sin \theta + k}{2} - 2 =0

I need one more condition so that I can eliminate theta.
 
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utkarshakash said:

Homework Statement


Find equation of the curve on reflection of the ellipse \dfrac{(x-4)^2}{16} + \dfrac{(y-3)^2}{9} = 1 about the line x-y-2=0.

Homework Equations



The Attempt at a Solution


Let the general point be P(4+4cosθ,3+3sinθ). Let the reflected point be (h,k).

\dfrac{4+4cos \theta + h}{2} - \dfrac{3+3sin \theta + k}{2} - 2 =0

I need one more condition so that I can eliminate theta.

Why not find the reflection of the point (4+4cosθ,3+3sinθ) about the given line? It gives h and k in terms of ##\theta## and its very straightforward from there.

Use the condition that line joining (h,k) and P is perpendicular to x-y-2=0. Can you take it from here?
 

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