Let L L prime be the latus rectum through the focus of the hyperbola (x2/a2)-(y2/b2)=1 and A prime be the farther vertex. If triangle A prime L L prime is equilateral and the eccentricity of the hyperbola (axes are coordinate axes) is (1+(1/ k )) then find k.

Question

Let L L prime be the latus rectum through the focus of the hyperbola (x2/a2)-(y2/b2)=1 and A prime be the farther vertex. If triangle A prime L L prime is equilateral and the eccentricity of the hyperbola (axes are coordinate axes) is (1+(1/ k )) then find k. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0fe3c2ee75d236314d4e77e23215be6f-.png" /> tan 30°=((b2/a)/a+a e) ⇒ (1+e/√3)=e2-1 ⇒ e -1=(1/√3) ⇒ e =(√3+1/√3)

Q. Let $L L^{'}$ be the latus rectum through the focus of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ and $A^{'}$ be the farther vertex. If $△ A^{'} L L^{'}$ is equilateral and the eccentricity of the hyperbola (axes are coordinate axes) is $(1 + \frac{1}{k})$ then find $k$ .

$tan 3 0^{\circ} = \frac{\frac{b ^{2}}{a}}{a + a e}$
$\Rightarrow \frac{1 + e}{3} = e^{2} - 1$
$\Rightarrow e - 1 = \frac{1}{3}$
$\Rightarrow e = \frac{3 + 1}{3}$

Q. Let LL′ be the latus rectum through the focus of the hyperbola a2x2​−b2y2​=1 and A′ be the farther vertex. If △A′LL′ is equilateral and the eccentricity of the hyperbola (axes are coordinate axes) is (1+k1​) then find k.

tan30∘=a+aeab2​​ ⇒3​1+e​=e2−1 ⇒e−1=3​1​ ⇒e=3​3​+1​

$tan 3 0^{\circ} = \frac{\frac{b ^{2}}{a}}{a + a e}$
$\Rightarrow \frac{1 + e}{3} = e^{2} - 1$
$\Rightarrow e - 1 = \frac{1}{3}$
$\Rightarrow e = \frac{3 + 1}{3}$