Find the equation of axis of the given hyperb | vedclass

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

Question

(a) $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$

Equation of tangent are equally inclined to the axis

$i.e.$, $	an 	heta = 1 = m$.

 Eq.

Vedclass · Accepted Answer

$y = x + 1$

Vedclass · Answer

(a) $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$

Equation of tangent are equally inclined to the axis

$i.e.$, $	an 	heta = 1 = m$.

 Eq. of tangent $y = mx + \sqrt {{a^2}{m^2} - {b^2}} $

Given eq. $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is a eq. of hyperbola

which is of form $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.

Now, on comparing ${a^2} = 3$, ${b^2} = 2$

 $y = 1.x + \sqrt {3 	imes {{(1)}^2} - 2} $

==> $y = x + 1$.

Find the equation of axis of the given hyperbola $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ which is equally inclined to the axes

Similar Questions

The differential equation $\frac{{dx}}{{dy}}= \frac{{3y}}{{2x}}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity :

The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} - 3{y^2} = 1$, is

The equation of the tangents to the hyperbola $4x^2 -y^2 = 12$ are $y = 4x+ c_1 \,$$ \& \, y = 4x + c_2,$ then $|c_1 -c_2|$ is equal to -

The latus-rectum of the hyperbola $16{x^2} - 9{y^2} = $ $144$, is

If the latus rectum of an hyperbola be 8 and eccentricity be $3/\sqrt 5 $, then the equation of the hyperbola is