Q. 60 - Calculus

Question

In Exercises 60 and 61 we ask you to prove Theorem 9.23 for ellipses and hyperbolas 

Consider the ellipse with equation $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ where $A>B$. Let $F$ be the focus with coordinates $\left(\sqrt{A^2-B^2},0\right)$. Let $e=\frac{\sqrt{A^2-B^2}}{A}$ and l be the vertical line with equation $x=\frac{A}{e}$ . Show that for any point P on the ellipse, $\frac{FP}{DP}=e_{\text{'}}$, where $D$is the point on $l$ closest to $P$. 

Step-by-Step Solution

Answer

Hence proved.

Step 1. Given information.

It is given,

$\frac{FP}{DP}=e^{\text{'}}$

$$\begin{gathered} F\text{ be the focus with coordinates }\left(\sqrt{A^2-B^2},0\right) \\ \mathit{e}\mathbf{=}\frac{\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{B}}^{\mathbf{2}}}}{\mathbf{A}} \\ \mathit{x}\mathbf{=}\frac{\mathbf{A}}{\mathbf{e}} \end{gathered}$$

Step 2. Explanation.

Using the distance formula,

$$\begin{gathered} FP=\sqrt{{\left(x-\left(\sqrt{A^2-B^2}\right)\right)}^2+(y-0{)}^2}\left[\text{ since }{x}_1=x,y_1=y,x_2=\sqrt{A^2-B^2},y_2=0\right] \\ FP=\sqrt{\left(x^2+{\left(\sqrt{A^2-B^2}\right)}^2-2x\sqrt{A^2-B^2}\right)+y^2} \\ FP=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)-B^2-\frac{B^2{x}^2}{A^2}}\left[since\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\Rightarrow y^2=B^2-\frac{B^2{x}^2}{A^2}\right] \\ FP=\sqrt{x^2+A^2-2x\sqrt{A^2-B^2}-\frac{B^2{x}^2}{A^2}} \\ FP=\sqrt{x^2\left(1-\frac{B^2}{A^2}\right)+A^2-2x\sqrt{A^2-B^2}} \\ FP=\sqrt{\left(\frac{A^2-B^2}{A^2}\right)\left(x^2-2x\sqrt{A^2-B^2}\frac{A^2}{A^2-B^2}+A^2·\frac{A^2}{A^2-B^2}\right)} \\ FP=\sqrt{\frac{A^2-B^2}{A^2}}\sqrt{\left(x^2-2x\sqrt{A^2-B^2}\frac{A^2}{{\left(\sqrt{A^2-B^2}\right)}^2}+A^2·\frac{A^2}{{\left(\sqrt{A^2-B^2}\right)}^2}\right)} \\ FP=\sqrt{\frac{A^2-B^2}{A^2}}\sqrt{{\left(x-\frac{A^2}{\sqrt{A^2-B^2}}\right)}^2} \\ FP=e\left(x-\frac{A}{e}\right)\left[since\frac{\sqrt{A^2-B^2}}{A}=e\right] \\ \frac{FP}{DP}=\frac{e\left(x-\frac{A}{e}\right)}{x-\frac{A}{e}}\left[\text{ since }FP=e\left(x-\frac{A}{e}\right),DP=x-\frac{A}{e}\right] \\ \frac{FP}{DP}=e \end{gathered}$$uncaught exception: Invalid chunk

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$$\begin{gathered} FP=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)+y^2} \\ FP=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)-B^2-\frac{B^2{x}^2}{A^2}}\left[\text{ since }\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\Rightarrow y^2=B^2-\frac{B^2{x}^2}{A^2}\right] \\ FP=\sqrt{x^2\left(1-\frac{B^2}{A^2}\right)+A^2-2x\sqrt{A^2-B^2}} \end{gathered}$$

$$\begin{gathered} FP=\sqrt{\left(\frac{A^2-B^2}{A^2}\right)\left(x^2-2x\sqrt{A^2-B^2}\frac{A^2}{A^2-B^2}+A^2·\frac{A^2}{A^2-B^2}\right)} \\ FP=\sqrt{\frac{A^2-B^2}{A^2}}\sqrt{\left(x^2-2x\sqrt{A^2-B^2}\frac{A^2}{{\left(\sqrt{A^2-B^2}\right)}^2}+A^2·\frac{A^2}{{\left(\sqrt{A^2-B^2}\right)}^2}\right)} \\ FP=\sqrt{\frac{A^2-B^2}{A^2}}\sqrt{\left(x^2-2x\frac{A^2}{\left(\sqrt{A^2-B^2}\right)}+{\left(\frac{A^2}{\left(\sqrt{A^2-B^2}\right)}\right)}^2\right)} \end{gathered}$$

$$\begin{gathered} FP=\sqrt{\frac{A^2-B^2}{A^2}}\left(x-\frac{A^2}{\sqrt{A^2-B^2}}\right) \\ FP=e\left(x-\frac{A}{e}\right)\left[since\frac{\sqrt{A^2-B^2}}{A}=e\right] \\ \frac{FP}{DP}=\frac{e\left(x-\frac{A}{e}\right)}{x-\frac{A}{e}}\left[\text{ since }FP=e\left(x-\frac{A}{e}\right),DP=x-\frac{A}{e}\right] \\ \frac{FP}{DP}=e \end{gathered}$$