For different values of alpha, the locus of t | vedclass

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

Question

(A) Given equations are:$L_1: \sqrt{3} x - y = 4 \sqrt{3} \alpha$  $(1)$$L_2: \alpha(\sqrt{3} x + y) = 4 \sqrt{3} \Rightarrow \sqrt{3} x + y = \frac{4

Vedclass · Accepted Answer

a hyperbola with eccentricity $2$

Vedclass · Answer

(A) Given equations are:$L_1: \sqrt{3} x - y = 4 \sqrt{3} \alpha$  $(1)$$L_2: \alpha(\sqrt{3} x + y) = 4 \sqrt{3} \Rightarrow \sqrt{3} x + y = \frac{4 \sqrt{3}}{\alpha}$  $(2)$Multiplying equation $(1)$ and $(2)$, we get:$(\sqrt{3} x - y)(\sqrt{3} x + y) = (4 \sqrt{3} \alpha) 	imes \left(\frac{4 \sqrt{3}}{\alpha}ight)$$3x^2 - y^2 = 16 	imes 3 = 48$Dividing by $48$, we get:$\frac{x^2}{16} - \frac{y^2}{48} = 1$This is the equation of a hyperbola where $a^2 = 16$ and $b^2 = 48$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$.Thus, it is a hyperbola with eccentricity $2$.

For different values of $\alpha$, the locus of the point of intersection of the two straight lines $\sqrt{3} x - y - 4 \sqrt{3} \alpha = 0$ and $\sqrt{3} \alpha x + \alpha y - 4 \sqrt{3} = 0$ is

Similar Questions

The eccentricity of a rectangular hyperbola is

If a circle intersects the hyperbola $xy = 1$ at four points $(x_r, y_r)$ for $r = 1, 2, 3, 4$,then:

Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $(0, \pm \frac{\sqrt{11}}{2})$.

If $3 \sqrt{2} x - 4 y = 12$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{5}{4}$ is its eccentricity,then $a^2 - b^2 =$

The number of common tangents that can be drawn to the curves $\frac{x^2}{16}-\frac{y^2}{9}=1$ and $x^2+y^2=16$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module