Download App

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

Maharashtra 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 $\text{b}^2=\text{a}^2(\text{e}^2-1).$
✓

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(5 Marks)" from Brief Review Of Cartesian System Of Rectangular Coordinates→Brief Review Of Cartesian System Of Rectangular Coordinates — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}+4\sin3\text{x}}{4\sin2\text{x}+7\text{x}}$
→
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
→
Find the sum of the series whose $n^{th}$ term is:
$2n^3 + 3n^2 - 1$
→
In $\triangle A B C$ prove that
$\frac{\cos A-\cos B+\cos C+1}{\cos A+\cos B+\cos C-1}=\cot \frac{A}{2} \cot \frac{C}{2}$
→
A bag contains 6 red, 4 white and 8 blue balls. if three balls are drewn at random, find the probability that:
(i) One is red and two are white
(ii) Two are blue and one is red
(iii) One is red.
→
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$1-\sin\alpha+\text{i}\cos\alpha$
→
Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3x + y = 12 which is intercepted between the axes of coordinates.
→
Represent to solution set of each of the following inequations graphically in two dimensional plane:
$3\text{x}-2\text{y}\leq\text{x}+\text{y}-8$
→
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
→
Find the $20^{th}$ term and the sum of $20$ terms of the series:
$2 × 4 + 4 × 6 + 6 × 8 + ....$
→