parametric trig curve

[red and white kop!]

find the equation of the normal to the curve x= 2cosALPHA, y = 3sinALPHA at the point where ALPHA= pi/4
find the coordinates of the point where this normal cuts the curve again.

i didnt have any problems with the first question; i found the equation of the normal is y = 2x/3 - 2sqrt2/3 + 3sqrt2/2
but from here its hard to find the point where the curve and the normal intersect again. i tried using 3sinALPHA=2x/3 - 2sqrt2/3 + 3sqrt2/2 and replacing ALPHA with arccos(x/2) but that doesnt make sense does it?

 

[BigGlenntheHeavy]

\(\displaystyle x \ = \ 2cos(\alpha), \ y \ = \ 3sin(\alpha), \ x \ = \ 2cos(\pi/4) \ = \ \sqrt2, \ y \ = \ 3sin(\pi/4) \ = \ 3\ \sqrt2/2\)

\(\displaystyle \frac{x^{2}}{2^{2}}+\frac{y^{2}}{3^{2}} \ =1, \ an \ ellipse, \ y^{2} \ = \ \frac{9}{4}(4-x^{2})\)

\(\displaystyle y \ = \ f(x) \ = \ \pm \frac{3}{2}\sqrt (4-x^{2}), \ y' \ = \ f '(x) \ = \ \frac{-3x}{2\sqrt(4-x^{2})}\)

\(\displaystyle f'(\sqrt2) \ = \ \frac{-3}{2} \ = \ m, \ hence \ equation \ tangent \ to \ the \ ellipse \ at \ (\sqrt2,3\sqrt2/2) \ = \ y \ = \ \frac{-3x}{2}+3\sqrt2.\)

\(\displaystyle Line \ normal \ to \ the \ ellipse \ at \ this \ point \ is \ y \ = \ \frac{2x}{3}+\frac{5\sqrt2}{6}\)

\(\displaystyle Normal \ line \ intercepts \ other \ side \ of \ ellipse \ at \ point \ (-1.99739,-.153085), \ see \ graph.\)

[attachment=0:2tdx9kpp]def.jpg[/attachment:2tdx9kpp]

Note: I've edited a few arithmetric errors, sorry, never did learn how to count.

 

[Deleted member 4993]

Equation of ellipse:

9x[sup:265f41uv]2[/sup:265f41uv] + 4y[sup:265f41uv]2[/sup:265f41uv] = 36 .........................(1)

equation of the normal is:

y = 2x/3 + 5*sqrt(2)/6............................................(2)

Replace 'y' in (1) with "2x/3 + 5*sqrt(2)/6" - and solve.