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Brief Review Of Cartesian System Of Rectangular Coordinates — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsBrief Review Of Cartesian System Of Rectangular Coordinates5 Marks
Question
A point moves as so that the difference its distances from (ae, 0) and (-ae, 0) is 2 a, prove that the equation to its locus is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1 $, where $\text{b}^2=\text{a}^2(\text{e}^2-1).$
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Answer

Let P(h, k) be any point on the locus and let A(ae, 0) and B(-ae, 0) be the given points. By the given condition$\text{PA}-\text{PB}=2\text{a}$
$\Rightarrow \text{PA}=2\text{a}+\text{PB}$
$\Rightarrow \sqrt{(\text{ae}-\text{h})^2+(0-\text{k})^2}=2\text{a}+\sqrt{(-\text{ae}-\text{h})^2+(0-\text{k})^2}$
$\Rightarrow (\text{ae}-\text{h})^2+\text{k}^2= \big(2\text{a}+\sqrt{(\text{ae}-\text{h}^2)+\text{k}^2}$ [Taking square on both sides]
$\Rightarrow \text{(ae)}^2+\text{h}^2-2\text{aeh}+\text{k}^2 =4\text{a}^2+(\text{a}\text{e}+\text{h})^2\\\ \ \ +\text{k}^2+2\times2\text{a}\times\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow \text{h}^2+\text{k}^2+\text{(ae)}^2-2\text{aeh} =4\text{a}^2+(\text{a}\text{e})^2+\text{h}^2\\\ \ \ +2\text{h}\text{a}\text{e}+\text{k}^2+4\text{a}\times\sqrt{(\text{a}\text{e}^2+\text{h}^2)+\text{k}^2}$
$\Rightarrow -4\text{a}-2\text{a}\text{e}\text{h}-2\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\text{a}^2-4\text{a}\text{e}\text{h}=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -4\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=4\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\big[\text{a}^2+\text{a}\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$
$\Rightarrow -\text{a}[\text{a}+\text{e}\text{h}\big]=\text{a}\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow -[\text{a}+\text{e}\text{h}\big]=\sqrt{(\text{a}\text{e}+\text{h})^2+\text{k}^2}$ $\Rightarrow (\text{a}+\text{e}\text{h})^2=\bigg(\sqrt{(\text{a}\text{e}+\text{h}^2)+\text{k}^2}\bigg)^2$ [Taking square on both sides] $\Rightarrow \text{a}^2+(\text{e}\text{h})^22\text{h}\text{a}\text{e}=(\text{a}\text{e}+\text{h})^2+\text{k}^2$ $\Rightarrow \text{a}^2+(\text{e}\text{h})^2+2\text{h}\text{a}\text{e}=(\text{a}\text{e})+\text{h}^2+2\text{h}\text{a}\text{e}+\text{k}^2$ $\Rightarrow \text{a}^2+\text{e}^2\text{h}^2=\text{a}^2\text{e}^2+\text{h}^2+\text{k}^2$ $\Rightarrow \text{e}^2\text{h}^2-\text{h}^2-\text{k}^2=\text{a}^2\text{e}^2-\text{a}^2$ $\Rightarrow \text{h}^2(\text{e}^2-1)-\text{k}^2=\text{a}^2(\text{e}^2-1)$ $\Rightarrow \frac{\text{h}^2(\text{e}^2-1)}{\text{a}^2(\text{e}^2-1)}-\frac{\text{k}^2}{\text{a}^2(\text{e}^2-1)}=1$ $\Rightarrow \frac{\text{h}^2}{\text{a}^2}-\frac{\text{k}^2}{\text{b}^2},\text{Where }\text{b}^2=\text{a}^2(\text{e}^2-1)$ $\therefore$ The locus of (h, k) is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1 $ Hence proved.

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