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2015-01-27 Trig in Ellipse.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage{graphicx} \usepackage{cancel} \usepackage{wrapfig} \usepackage{txfonts} \usepackage[hmargin=2cm,vmargin=2cm]{geometry} \parindent0em \setlength{\parskip}{0.3cm} \begin{document} \hspace*{-0.5cm}{\large Q7} $P(x_1,y_1)$ is any point on the ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. a) Find the equation of the tangent at $P$. Ans: $\dfrac{x_1x}{a^2}+\dfrac{y_1y}{b^2}=1$. b) A line is draw from the centre (0,0) parallel to the tangent at $P$, meets the ellipse at $Q$. Prove that the area of triangle $OPQ$ is independent of the position of $P$. The gradient of the tangent is\quad $\dfrac{y}{x}=-\dfrac{b^2x_1}{a^2y_1}$. Let us find $Q(x,y)$ by subsituting the gradient into the ellipse equation:\quad $b^2x^2+a^2y^2=a^2b^2$. $b^2x^2+a^2\left(-\dfrac{b^2x_1}{a^2y_1}\cdot x\right)^2=a^2b^2$,\qquad $\cancel{b^2}\left(1+\dfrac{b^2x_1^2}{a^2y_1^2}\right)x^2=a^2\cancel{b^2}$,\qquad $x^2=\dfrac{a^4y_1^2}{a^2y_1^2+b^2x_1^2}=\dfrac{a^4y_1^2}{a^2b^2}$. Subsitute $x^2$ back to the ellipse for $y^2$: $b^2\cdot\dfrac{a^{\cancel{4}~2}y_1^2}{a^2y_1^2+b^2x_1^2}+\cancel{a^2}y^2=\cancel{a^2}b^2$,\qquad $y^2=b^2\left(1-\dfrac{a^2y_1^2}{a^2y_1^2+b^2x_1^2}\right)$,\qquad $y^2=\dfrac{b^4x_1^2}{a^2y_1^2+b^2x_1^2}=\dfrac{b^4x_1^2}{a^2b^2}$. $OQ$:\quad$\sqrt{x^2+y^2} =\sqrt{\dfrac{a^4y_1^2+b^4x_1^2}{a^2b^2}} =\dfrac{\sqrt{a^4y_1^2+b^4x_1^2}}{ab}$. The distance from 0 to $P$:\quad $d_0=\left|\dfrac{1}{\sqrt{\left(\dfrac{x_1}{a^2}\right)^2+\left(\dfrac{y_1}{b^2}\right)^2}}\right| =\dfrac{a^2b^2}{\sqrt{b^4x_1^2+a^4y_1^2}}$ Area of triangle $OPQ =\frac{1}{2}\cdot OQ\cdot d_0 =\frac{1}{2}\cdot\dfrac{\cancel{\sqrt{a^4y_1^2+b^4x_1^2}}}{ab}\cdot\dfrac{a^2b^2}{\cancel{\sqrt{b^4x_1^2+a^4y_1^2}}} =\frac{1}{2}~ab. $ \end{document}