Ellipse perpendicular distance

[Rikendogenz]

Homework Statement

Prove that the straight line x cos @ + y sin @ = p is a tangent to the ellipse x²/a² + y²/b² if a² cos²@ + b² sin²@ =p² .
u and v are the perpendicular distances of a tangent from the two points M(0,ae) and N(0,-ae) respectively. Prove that u² + v² is a constant.


2. The attempt at a solution

So far I have: xx₁/a²+yy₁/b² -1 = x cos@ + y sin@ - p
I don't know where to go from here and I know in the second part I have to use the perpendicular distance formula but I'm not sure what to sub in and why.

 

[HallsofIvy]

Homework Statement

Prove that the straight line x cos @ + y sin @ = p is a tangent to the ellipse x²/a² + y²/b² if a² cos²@ + b² sin²@ =p² .
u and v are the perpendicular distances of a tangent from the two points M(0,ae) and N(0,-ae) respectively. Prove that u² + v² is a constant.


2. The attempt at a solution

So far I have: xx₁/a²+yy₁/b² -1 = x cos@ + y sin@ - p
I don't know where to go from here and I know in the second part I have to use the perpendicular distance formula but I'm not sure what to sub in and why.

You mean the elliplse x^2/a^2+ y^2/b^2=1. You can simplify that a little by writing it as b^2x^2+ a^2y^2= a^2b^2. Differentiating with respect to x, 2b^2x+ 2a^2yy'= 0 so y'= -b^2x/a^2y. The tangent line at the point (x_1, y_1) is given by y= -(b^2x_1/a^2y_1)(x- x_1)+ y_1 or b^2x_1x+ a^2y_1y= b^2x_1^2+ a^2y_1^2[/tex].

If we are going to be able to write the coefficients as cos(\theta)[/itex] and sin(\theta), we must We must have cos^2(\theta)+ sin^2(\theta)= 1. Now we have b^4x_1^2+ a^4y_1^2 as the sum of the squares of the coefficients so divide the entire equation by the square root of that:<br /> \frac{b^2x_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}x+ \frac{a^2y_1}{\sqrt{b^4x_1^2+ a^4y_1^2}}y= \frac{b^2x_1^2+ a^2y_1^2}{\sqrt{b^4x_1^2+ a^4y_1^2}}<br /> <br /> That is of the form cos(\theta)x+ sin(\theta)b= p