If (l, m) is the circumcentre of an equilater | vedclass

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Question

(C) For an equilateral triangle,the circumcentre coincides with the centroid. The coordinates of the vertices are $(a \cos 	heta_i, b \sin 	heta_i)$

Vedclass · Accepted Answer

$\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1$

Vedclass · Answer

(C) For an equilateral triangle,the circumcentre coincides with the centroid. The coordinates of the vertices are $(a \cos 	heta_i, b \sin 	heta_i)$ for $i=1, 2, 3$. Thus,$(l, m) = \left(\frac{a(\cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3)}{3}, \frac{b(\sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3)}{3}ight)$. This gives $\frac{3l}{a} = \cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3$ and $\frac{3m}{b} = \sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3$. Squaring and adding these equations: $\frac{9l^2}{a^2} + \frac{9m^2}{b^2} = (\cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3)^2 + (\sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3)^2$. Expanding the squares: $\frac{9l^2}{a^2} + \frac{9m^2}{b^2} = 3 + 2(\cos 	heta_1 \cos 	heta_2 + \cos 	heta_2 \cos 	heta_3 + \cos 	heta_3 \cos 	heta_1 + \sin 	heta_1 \sin 	heta_2 + \sin 	heta_2 \sin 	heta_3 + \sin 	heta_3 \sin 	heta_1)$. Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $\frac{9l^2}{a^2} + \frac{9m^2}{b^2} = 3 + 2[\cos(	heta_1-	heta_2) + \cos(	heta_2-	heta_3) + \cos(	heta_3-	heta_1)]$. Dividing by $3$: $\frac{3l^2}{a^2} + \frac{3m^2}{b^2} = 1 + \frac{2}{3}[\cos(	heta_1-	heta_2) + \cos(	heta_2-	heta_3) + \cos(	heta_3-	heta_1)]$. Therefore,$\frac{2}{3}[\cos(	heta_1-	heta_2) + \cos(	heta_2-	heta_3) + \cos(	heta_3-	heta_1)] = \frac{3l^2}{a^2} + \frac{3m^2}{b^2} - 1$.

List-$I$	List-$II$
$(P)$ Directrix corresponding to the focus $(-3, 0)$	$(1)$ $y = 4$
$(Q)$ Tangent at the vertex $(0, 4)$	$(2)$ $3x = 25$
$(R)$ Latus rectum through $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

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