If for theta in left[-frac{pi}{3}, 0right],th | vedclass

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

Question

(D) Given $x = 3 	an \left(	heta + \frac{\pi}{3}ight) = 3 \left( \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} ight)$.Rearranging,$x(1

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) Given $x = 3 	an \left(	heta + \frac{\pi}{3}ight) = 3 \left( \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} ight)$.Rearranging,$x(1 - \sqrt{3} 	an 	heta) = 3 	an 	heta + 3\sqrt{3}$ $\Rightarrow x - 3\sqrt{3} = 	an 	heta (3 + \sqrt{3}x)$ $\Rightarrow 	an 	heta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} \dots (1)$.Given $y = 2 	an \left(	heta + \frac{\pi}{6}ight) = 2 \left( \frac{	an 	heta + \frac{1}{\sqrt{3}}}{1 - \frac{	an 	heta}{\sqrt{3}}} ight) = 2 \left( \frac{\sqrt{3} 	an 	heta + 1}{\sqrt{3} - 	an 	heta} ight)$.Rearranging,$y(\sqrt{3} - 	an 	heta) = 2\sqrt{3} 	an 	heta + 2 \dots (2)$.Substituting $(1)$ into $(2)$:$y \left( \sqrt{3} - \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} ight) = 2\sqrt{3} \left( \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} ight) + 2$.$y \left( \frac{3\sqrt{3} + 3x - x + 3\sqrt{3}}{3 + \sqrt{3}x} ight) = \frac{2\sqrt{3}x - 18 + 6 + 2\sqrt{3}x}{3 + \sqrt{3}x}$.$y(2x + 6\sqrt{3}) = 4\sqrt{3}x - 12 \Rightarrow 2xy + 6\sqrt{3}y = 4\sqrt{3}x - 12$.Dividing by $2$: $xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0$.Comparing with $xy + \alpha x + \beta y + \gamma = 0$,we get $\alpha = -2\sqrt{3}$,$\beta = 3\sqrt{3}$,$\gamma = 6$.Thus,$\alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + (6)^2 = 12 + 27 + 36 = 75$.

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

Similar Questions

Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is:

Identify the statements which are True.

The locus of the foot of the perpendicular from the centre of the hyperbola $xy = c^2$ on a variable tangent is :

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. The equation of the common tangent with a positive slope to the circle and the hyperbola is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module