If a directrix of a hyperbola centered at the | vedclass

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

Question

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2 \sqrt{3})$,we have $\frac{16}{a^2} - \fr

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2 \sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} = 1$. Using $b^2 = a^2(e^2 - 1)$,we get $\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$,which simplifies to $16(e^2 - 1) - 12 = a^2(e^2 - 1) \Rightarrow 16e^2 - 28 = a^2(e^2 - 1) \quad (i)$. The directrix is $x = \frac{a}{e}$,so $\frac{a}{e} = \frac{4}{\sqrt{5}} \Rightarrow a^2 = \frac{16e^2}{5} \quad (ii)$. Substituting $(ii)$ into $(i)$: $16e^2 - 28 = \frac{16e^2}{5}(e^2 - 1)$. Multiplying by $5$: $80e^2 - 140 = 16e^4 - 16e^2 \Rightarrow 16e^4 - 96e^2 + 140 = 0$. Dividing by $4$: $4e^4 - 24e^2 + 35 = 0$. Factoring the quadratic in $e^2$: $(2e^2 - 7)(2e^2 - 5) = 0$. Thus,$e^2 = \frac{7}{2}$ or $e^2 = \frac{5}{2}$. Given the options,$e^2 = \frac{7}{2}$ is the correct value.

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2 \sqrt{3})$ is $\sqrt{5}x = 4$ and $e$ is its eccentricity,then $e^2 =$

Similar Questions

The length of the transverse axis of the hyperbola $3x^2 - 4y^2 = 32$ is

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

Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^{2}x^{2} - y^{2} = b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2} - \left(\frac{\mu}{b}\right)^{2}$ is equal to

If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$ respectively,then $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to

The length of the latus rectum of the hyperbola $25x^2 - 16y^2 = 400$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module