The values of m for which the line y=mx+2 is | vedclass

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

Question

(B) The equation of the line is $y=mx+2$ ... $(i)$ The equation of the hyperbola is $4x^2-9y^2=36$ ... $(ii)$ Dividing $(ii)$ by $36$,we get $\frac{x^

Vedclass · Accepted Answer

$\pm \frac{2\sqrt{2}}{3}$

Vedclass · Answer

(B) The equation of the line is $y=mx+2$ ... $(i)$ The equation of the hyperbola is $4x^2-9y^2=36$ ... $(ii)$ Dividing $(ii)$ by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. For a line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition is $c^2 = a^2m^2 - b^2$. Here,$a^2=9$,$b^2=4$,and $c=2$. Substituting these values into the condition: $2^2 = 9m^2 - 4$ $4 = 9m^2 - 4$ $9m^2 = 8$ $m^2 = \frac{8}{9}$ $m = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}$

The values of $m$ for which the line $y=mx+2$ is a tangent to the hyperbola $4x^2-9y^2=36$ are

Similar Questions

If $A(4,0)$ and $B(-4,0)$ are two points,then the locus of a point $P$ such that $PA - PB = 4$ is

If the circle $x^2 + y^2 = a^2$ intersects the hyperbola $xy = c^2$ in four points $P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4)$,then:

The distance between the tangents of the hyperbola $2x^2 - 3y^2 = 6$ which are perpendicular to the line $x - 2y + 5 = 0$ is

The foci of a hyperbola are $(\pm 3, 0)$ and the equation of a tangent is $2x + y - 4 = 0$. Find the equation of the hyperbola.

The angle between the tangents drawn from $(1, 2\sqrt{2})$ to the hyperbola $16x^{2} - 25y^{2} = 400$ is.....

Vedclass Test Series

Exam Paper Generator

Online Exam Module