Question

The locus of the centroid of the triangle with vertices at (acos teeta, asin teeta), (bsin teeta, bcos teeta) and (1,0) is (Here is a parameter)

1) (3x - 1)^2 +9y^2 = a^2 + b^2
2) (3x + 1)^2 +9y^2 = a^2 + b^2
3) (3x + 1)^2 + 9y^2= a^2 + b^2
4) (3x - 1)^2 +9y^2= a^2 - b^2​

Answer 1

Answer:

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Step-by-step explanation:

$so \: let \: \\ A = (x1,y1) = (a.cos \: θ,a \: sin \: θ) \\ B =(x2 , y2) = (b. sin \: θ , b.cos \: θ) \\ C = (x3,y3) = (1,0)$

$so \: by \: using \: centroid \: formula \\ we \: get \\ x = \frac{x1 + x2 + x3}{3} \: \: \: \: y = \frac{y1 + y2 + y3}{3} \\ \\ ie \: \: x = \frac{a.cos \:θ + b.sin \:θ + 1 }{3} \\ \\ 3x = a.cos \: θ + b.sin \: θ + 1 \\ \\ \: a.cos \: θ + b.sin \: θ = 3x - 1 \: \: \: \: \: \: \: \: \: (1)$

$similarly \: using \\ \\ y = \frac{y1 + y2 + y3}{3} \\ \\ = \frac{a.sin \:θ + b.cos \: θ + 0}{3} \\ \\ ie \: \: 3y = a.sin \: θ + b.cos \: θ \: \: \: \: \: \: \: \: \: \: (2)$

$now \: squaring \: and \: adding \: both \: equations \\ we \: get \\ (3x - 1) {}^{2} + (3y) {}^{2} = (a.cos \: θ + b.sin \: θ) {}^{2} + (a.sin \: θ + b.cos \: θ) {}^{2} \\ \\ 9x {}^{2} + 1 - 6x + 9y {}^{2} = a {}^{2} .cos {}^{2} \: θ + b {}^{2} .sin {}^{2} \: θ + 2ab.sin \: θ.cos \: θ + a {}^{2} .sin {}^{2} θ + b {}^{2} .cos {}^{2} θ + 2ab.sin \: θ.cos \: θ\\ \\ 9x {}^{2} + 1 + 9y {}^{2} - 6x = a {}^{2} .cos {}^{2} \: θ + a {}^{2} .sin {}^{2} θ + b {}^{2} .sin {}^{2} \: θ + b {}^{2} .cos {}^{2} \: θ + 4ab.sin \: θ.cos \: θ \\ \\ = a {}^{2} \: (sin {}^{2} θ + cos { }^{2} θ) + b {}^{2} (sin {}^{2} θ + cos {}^{2} θ) + 4ab.sin \: θ.cos \: θ \\ \\ = a {}^{2} (1) + b {}^{2} (1) + 4ab.sin \: θ.cos \: θ$

$9x {}^{2} + 9y {}^{2} + 1 - 6x = a {}^{2} + b {}^{2} + 4ab.sin \: θ.cos \: θ \\ is \: required \: equation \: of \: locus \: of \: centroid$