The equation of a line passing through the ce | vedclass

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

Question

(D) For a rectangular hyperbola,the asymptotes are mutually perpendicular. Given one asymptote is $3x - 4y - 6 = 0$,the other must be of the form $4x

Vedclass · Accepted Answer

$4x + 3y + 17 = 0$

Vedclass · Answer

(D) For a rectangular hyperbola,the asymptotes are mutually perpendicular. Given one asymptote is $3x - 4y - 6 = 0$,the other must be of the form $4x + 3y + \lambda = 0$.The centre of the hyperbola is the intersection point of the two asymptotes.Solving $3x - 4y - 6 = 0$ and $4x + 3y + \lambda = 0$:Multiply the first by $3$ and the second by $4$: $9x - 12y - 18 = 0$ and $16x + 12y + 4\lambda = 0$.Adding these gives $25x + 4\lambda - 18 = 0$,so $x = \frac{18 - 4\lambda}{25}$.Substituting $x$ into $3x - 4y - 6 = 0$: $4y = 3(\frac{18 - 4\lambda}{25}) - 6 = \frac{54 - 12\lambda - 150}{25} = \frac{-96 - 12\lambda}{25}$,so $y = \frac{-24 - 3\lambda}{25}$.Since the centre lies on $x - y - 1 = 0$,we have $\frac{18 - 4\lambda}{25} - (\frac{-24 - 3\lambda}{25}) - 1 = 0$.$18 - 4\lambda + 24 + 3\lambda - 25 = 0$ $\Rightarrow 17 - \lambda = 0$ $\Rightarrow \lambda = 17$.Thus,the equation of the other asymptote is $4x + 3y + 17 = 0$.

The equation of a line passing through the centre of a rectangular hyperbola is $x - y - 1 = 0$. If one of the asymptotes is $3x - 4y - 6 = 0$,the equation of the other asymptote is:

Similar Questions

If the eccentricity of a hyperbola is $\sqrt{3}$,then the eccentricity of its conjugate hyperbola is:

From any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area of the figure formed by the chord of contact of that point and the asymptotes is

The equation of the hyperbola referred to its axes as axes of coordinate and whose distance between the foci is $16$ and eccentricity is $\sqrt{2}$ is:

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

If the line $ax + by = 1$ is normal to the hyperbola $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$,then $\frac{p^2}{a^2} - \frac{q^2}{b^2}$ is equal to (where $a, b, p, q \in R^+$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module