The locus of a point of intersection of two l | vedclass

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

Question

(B) Given the equations of the two lines: $x \sqrt{3}-y=k \sqrt{3} \quad ...(1)$ $\sqrt{3} k x+k y=\sqrt{3} \quad ...(2)$ From equation $(1)$,we have

Vedclass · Accepted Answer

a hyperbola

Vedclass · Answer

(B) Given the equations of the two lines: $x \sqrt{3}-y=k \sqrt{3} \quad ...(1)$ $\sqrt{3} k x+k y=\sqrt{3} \quad ...(2)$ From equation $(1)$,we have $k = \frac{x \sqrt{3}-y}{\sqrt{3}}$. From equation $(2)$,we have $k = \frac{\sqrt{3}}{\sqrt{3} x+y}$. Equating the values of $k$ from both equations: $\frac{x \sqrt{3}-y}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3} x+y}$ $(x \sqrt{3}-y)(\sqrt{3} x+y) = (\sqrt{3})(\sqrt{3})$ $3x^{2} - y^{2} = 3$ Dividing by $3$,we get $\frac{x^{2}}{1} - \frac{y^{2}}{3} = 1$. This is the standard equation of a hyperbola,$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.

The locus of a point of intersection of two lines $x \sqrt{3}-y=k \sqrt{3}$ and $\sqrt{3} k x+k y=\sqrt{3}$,where $k \in R$,describes:

Similar Questions

Find the eccentricity of the conic represented by $x^{2} - y^{2} - 4x + 4y + 16 = 0$.

If one focus of a hyperbola is $(3,0)$,the equation of its directrix is $4x - 3y - 3 = 0$ and its eccentricity is $e = \frac{5}{4}$,then the coordinates of its vertex are:

The foci of the hyperbola $5x^2 - 6y^2 - 10x - 24y - 34 = 0$ are

The product of the lengths of the perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is

If the equation of a hyperbola is $9x^2 - 16y^2 + 72x - 32y - 16 = 0$,then the equation of its conjugate hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module