For different values of alpha,the locus of th | 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

If $\frac{(3x - 4y - z)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$,then the length of the latus rectum of the hyperbola is:

If the circle $x^{2}+y^{2}=a^{2}$ intersects the hyperbola $xy=c^{2}$ in four points $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3})$ and $S(x_{4}, y_{4})$,then

The equation of the hyperbola having its eccentricity $e = 2$ and the distance between its foci as $8$ is:

The equation of the chord of the hyperbola $25x^{2} - 16y^{2} = 400$ whose midpoint is $(5, 3)$ is:

The area of the triangle formed by the tangent to the hyperbola $xy = a^2$ and the coordinate axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module