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

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To find the equation of the curve reflected over the line x-y-2=0, the initial ellipse equation is given as (x-4)²/16 + (y-3)²/9 = 1. The general point on the ellipse is represented as P(4+4cosθ, 3+3sinθ), and the reflected point is denoted as (h,k). A key step involves establishing a relationship between P and (h,k) using the condition that the line connecting these points is perpendicular to the line x-y-2=0. The discussion emphasizes the need for an additional condition to eliminate θ for solving the reflection. The approach suggests finding the reflection of point P directly to derive 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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