Brief Review of Cartesian System of Rectangular Coordinates — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBrief Review of Cartesian System of Rectangular Coordinates3 Marks
Question
Find the equation of the locus of a point which moves such that the ratio of distances from (2, 0) and (1, 3) is 5 : 4.
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Answer
Let P(h, k) be any point on the locus and let A(2, 0) and B(1, 3). Then, $\frac{\text{PA}}{\text{BP}}=\frac{5}{4}$ $\frac{\text{PA}^2}{\text{BP}^2}=\frac{25}{16}$$\Rightarrow \frac{\bigg[\sqrt{(\text{h}-2)^2+(\text{k}-0)^2}\bigg]}{\bigg[\sqrt{(\text{h}-1)^2+(\text{k}-0)^3}\bigg]}=\frac{25}{16}$
$\Rightarrow \frac{(\text{h}-2)^2+\text{k}^2}{(\text{h}-1)^2+(\text{k}^3-3)^2}=\frac{25}{16}$
$\Rightarrow \frac{\text{h}^2+4-4\text{h}+\text{k}^2}{\text{h}^2+1-2\text{h}+\text{k}^2+9-6\text{k}}=\frac{25}{16}$
$\Rightarrow \frac{(\text{h}^2-4\text{h}+\text{k}^2+4)}{\text{h}^2+\text{k}^2-2\text{h}-6\text{k}+10}=\frac{25}{16}$
$\Rightarrow 16(\text{h}^2-4\text{h}+\text{k}^2+4)=25(\text{h}^2+\text{k}^2-2\text{h}-6\text{k}+10)$
$\Rightarrow 16\text{h}^2-64\text{h}+16\text{k}^2+64=25\text{h}^2+25\text{k}^2-50\text{h}-150\text{k}+250$
$\Rightarrow 25\text{h}^2-16\text{h}^2+25\text{k}^2-16\text{k}^2-50\text{h}+64\text{h}-150\text{k}+250-65=0$
$\Rightarrow 9\text{h}^2+9\text{k}^2+14\text{k}-150\text{k}+186=0$ Hence, locus of (h, k) is $ 9\text{h}^2+9\text{k}^2+14\text{k}-150\text{k}+186=0$
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