The equation of one of the tangents drawn fro | vedclass

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

Question

(D) Let the equation of the tangent passing through $(0, 1)$ with slope $m$ be $y - 1 = m(x - 0)$,which simplifies to $y = mx + 1$.Substituting this i

Vedclass · Accepted Answer

$9x - 2y + 2 = 0$

Vedclass · Answer

(D) Let the equation of the tangent passing through $(0, 1)$ with slope $m$ be $y - 1 = m(x - 0)$,which simplifies to $y = mx + 1$.Substituting this into the hyperbola equation $45x^2 - 4y^2 = 5$:$45x^2 - 4(mx + 1)^2 = 5$$45x^2 - 4(m^2x^2 + 2mx + 1) = 5$$(45 - 4m^2)x^2 - 8mx - 9 = 0$For the line to be a tangent,the discriminant $D$ must be $0$:$D = (-8m)^2 - 4(45 - 4m^2)(-9) = 0$$64m^2 + 36(45 - 4m^2) = 0$$64m^2 + 1620 - 144m^2 = 0$$-80m^2 + 1620 = 0$$80m^2 = 1620$$m^2 = \frac{1620}{80} = \frac{81}{4}$$m = \pm \frac{9}{2}$Substituting $m = \frac{9}{2}$ into $y = mx + 1$:$y = \frac{9}{2}x + 1$ $\Rightarrow 2y = 9x + 2$ $\Rightarrow 9x - 2y + 2 = 0$Substituting $m = -\frac{9}{2}$ into $y = mx + 1$:$y = -\frac{9}{2}x + 1$ $\Rightarrow 2y = -9x + 2$ $\Rightarrow 9x + 2y - 2 = 0$Thus,the correct option is $D$.

The equation of one of the tangents drawn from the point $(0, 1)$ to the hyperbola $45x^2 - 4y^2 = 5$ is

Similar Questions

The eccentricity of the conic ${x^2} - 4{y^2} = 1$ is

Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is

The transverse axis of a hyperbola is along the $x$-axis and its length is $2a$. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

The equation of the normal to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at $(-4, 0)$ is

Find the equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module