Let the eccentricity of the hyperbola frac{x^ | vedclass

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

Question

(B) Given eccentricity $e = \frac{5}{4}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}} = \frac{25}{16}$ $\Rightarrow \frac{b^{2}}{a^{2}} = \frac{9}{16}$ $\Right

Vedclass · Accepted Answer

$85$

Vedclass · Answer

(B) Given eccentricity $e = \frac{5}{4}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}} = \frac{25}{16}$ $\Rightarrow \frac{b^{2}}{a^{2}} = \frac{9}{16}$ $\Rightarrow b^{2} = \frac{9}{16}a^{2}$.The point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}ight)$ lies on the hyperbola,so $\frac{64}{5a^{2}} - \frac{144}{25b^{2}} = 1$.Substituting $b^{2} = \frac{9}{16}a^{2}$,we get $\frac{64}{5a^{2}} - \frac{144}{25 	imes (9/16)a^{2}} = 1$ $\Rightarrow \frac{64}{5a^{2}} - \frac{144 	imes 16}{225a^{2}} = 1$ $\Rightarrow \frac{64}{5a^{2}} - \frac{256}{25a^{2}} = 1$.$\frac{320 - 256}{25a^{2}} = 1$ $\Rightarrow \frac{64}{25a^{2}} = 1$ $\Rightarrow a^{2} = \frac{64}{25}$ $\Rightarrow a = \frac{8}{5}$.Then $b^{2} = \frac{9}{16} 	imes \frac{64}{25} = \frac{9 	imes 4}{25} = \frac{36}{25} \Rightarrow b = \frac{6}{5}$.The equation of the normal at $(x_{1}, y_{1})$ is $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$.Substituting values: $\frac{(64/25)x}{8/\sqrt{5}} + \frac{(36/25)y}{12/5} = \frac{64}{25} + \frac{36}{25} = \frac{100}{25} = 4$.$\frac{8\sqrt{5}}{25}x + \frac{3}{5}y = 4 \Rightarrow 8\sqrt{5}x + 15y = 100$.Comparing with $8\sqrt{5}x + \beta y = \lambda$,we get $\beta = 15$ and $\lambda = 100$.Thus,$\lambda - \beta = 100 - 15 = 85$.

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $8 \sqrt{5} x + \beta y = \lambda$,then $\lambda - \beta$ is equal to

Similar Questions

The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ form with any tangent to the hyperbola a triangle whose area is $a^2 \tan \lambda$ in magnitude. Then its eccentricity $e$ is:

The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 3$ is:

If one focus of a hyperbola is $(3,0)$,the equation of its directrix is $4x - 3y - 3 = 0$ and its eccentricity is $e = \frac{5}{4}$,then the coordinates of its vertex are:

Let the eccentricity of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and the length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$,then the value of $c^2$ is equal to

The equation of the tangents to the conic $3x^2 - y^2 = 3$ perpendicular to the line $x + 3y = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module