For t in (0, 2pi),if ABC is an equilateral tr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $s = \sin t$ and $c = \cos t$.For an equilateral triangle,the orthocentre $(h, k)$ coincides with the centroid.The centroid $(h, k)$ is given

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $s = \sin t$ and $c = \cos t$.For an equilateral triangle,the orthocentre $(h, k)$ coincides with the centroid.The centroid $(h, k)$ is given by $h = \frac{a + s + c}{3}$ and $k = \frac{b - c + s}{3}$.Thus,$3h - a = s + c$ and $3k - b = s - c$.Squaring and adding these equations:$(3h - a)^2 + (3k - b)^2 = (s + c)^2 + (s - c)^2 = 2(s^2 + c^2) = 2$.Dividing by $9$:$(h - a/3)^2 + (k - b/3)^2 = 2/9$.This represents a circle with centre $(a/3, b/3)$ and radius $\sqrt{2}/3$.Given the centre of the circle is $(1, 1/3)$,we have $a/3 = 1$ and $b/3 = 1/3$.Therefore,$a = 3$ and $b = 1$.Finally,$a^2 - b^2 = 3^2 - 1^2 = 9 - 1 = 8$.

For $t \in (0, 2\pi)$,if $ABC$ is an equilateral triangle with vertices $A(\sin t, -\cos t)$,$B(\cos t, \sin t)$,and $C(a, b)$ such that its orthocentre lies on a circle with centre $(1, 1/3)$,then $(a^2 - b^2)$ is equal to.

Similar Questions

Find the locus of a point such that its distance from the point $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is

The angle between a pair of tangents drawn from a point $P$ to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$. The equation of the locus of the point $P$ is

If $A(2, 3)$ and $B(2, -3)$ are two points,then the equation of the locus of a point $P$ such that $PA + PB = 8$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module