Normal to a fixed concentric ellipse

[utkarshakash]

Homework Statement

PM and PN are perpendiculars upon the axes from any point 'P' on the ellipse. Prove that MN is always normal to a fixed concentric ellipse

Homework Equations

The Attempt at a Solution

I assume point P to be (acosθ, bsinθ)

The eqn of line MN is then given by
$bsin\theta x+acos\theta y =absin\theta cos\theta$

 

[haruspex]

Try writing the generic equation for a normal to an ellipse with same centre and axes.

 

[utkarshakash]

Try writing the generic equation for a normal to an ellipse with same centre and axes.

$a'sec \phi x - b'cosec \phi y = a^2 - b^2$

 

[haruspex]

$a'sec \phi x - b'cosec \phi y = a^2 - b^2$

I assume you meant $a'sec \phi x - b'cosec \phi y = a'^2 - b'^2$
It remains to find expressions for a', b' and phi in terms of a, b and θ that make this the same as the equation for MN. There is a constraint regarding which of a', b' and phi can depend on which of a, b and θ.

 

[utkarshakash]

I assume you meant $a'sec \phi x - b'cosec \phi y = a'^2 - b'^2$
It remains to find expressions for a', b' and phi in terms of a, b and θ that make this the same as the equation for MN. There is a constraint regarding which of a', b' and phi can depend on which of a, b and θ.

Do you want me to compare the two lines?

 

[haruspex]

Do you want me to compare the two lines?

Yes. You want to make a′x sec(ϕ)−b′y cosec(ϕ)=a′²−b′² look like bx sin(θ)+ay cos(θ)=ab sinθ cosθ by suitable choices of a', b' and ϕ. But note that this must work keeping a' and b' fixed while ϕ is allowed to vary as a function of θ.

 

[utkarshakash]

Yes. You want to make a′x sec(ϕ)−b′y cosec(ϕ)=a′²−b′² look like bx sin(θ)+ay cos(θ)=ab sinθ cosθ by suitable choices of a', b' and ϕ. But note that this must work keeping a' and b' fixed while ϕ is allowed to vary as a function of θ.

OK I did exactly what you said and got the following relations after comparison

$cos\alpha = \dfrac{aa'cos\theta}{a'^2 - b'^2} \\<br /> <br /> sin \alpha = \dfrac{-bb'sin\theta}{a'^2 - b'^2} \\ <br /> <br /> tan \alpha = \dfrac{-bb' tan\theta}{aa'}<br />$

 

[haruspex]

OK I did exactly what you said and got the following relations after comparison

$cos\alpha = \dfrac{aa'cos\theta}{a'^2 - b'^2} \\<br /> <br /> sin \alpha = \dfrac{-bb'sin\theta}{a'^2 - b'^2} \\ <br /> <br /> tan \alpha = \dfrac{-bb' tan\theta}{aa'}<br />$

OK, nearly there. Now you must set a' and b' so that all values of θ and α can occur. (Otherwise there will be a gap in one of the ellipses.)