The equation of a tangent to the hyperbola 4x | vedclass

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

Question

(A) The given hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Comparing this with $\frac{x^2}{a^2} - \fra

Vedclass · Accepted Answer

$x - y + 1 = 0$

Vedclass · Answer

(A) The given hyperbola is $4x^2 - 5y^2 = 20$,which can be written as $\frac{x^2}{5} - \frac{y^2}{4} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 5$ and $b^2 = 4$. The given line is $x - y = 2$,which can be written as $y = x - 2$. The slope of this line is $m = 1$. The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting $m = 1$,$a^2 = 5$,and $b^2 = 4$,we get $y = 1(x) \pm \sqrt{5(1)^2 - 4} = x \pm \sqrt{5 - 4} = x \pm 1$. Thus,the tangents are $y = x + 1$ or $y = x - 1$,which can be written as $x - y + 1 = 0$ or $x - y - 1 = 0$. Comparing with the given options,$x - y + 1 = 0$ is the correct choice.

The equation of a tangent to the hyperbola $4x^2 - 5y^2 = 20$ parallel to the line $x - y = 2$ is

Similar Questions

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $\theta + \phi = \frac{\pi}{2}$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k = \dots$

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

The line $y = mx + c$ touches the curve $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,if

The equation of the normal to the curve $3x^2 - y^2 = 8$,which is parallel to the line $x + 3y = 10$,is

If $5x^2 + \lambda y^2 = 20$ represents a rectangular hyperbola,then $\lambda$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module