Question

A hyperbola, having the transverse axis of length $3x^2+4y^2=12$. Its equation is

• A. $x^2{\sin }^2\theta -y^2{\cos }^2\theta =1$
• B. $x^{2}co{\sec }^2\theta -y^2{\sec }^2\theta =1$
• C. $(x^2+y^2){\sin }^2\theta =1+y^2$
• D. $x^{2}co{\sec }^2\theta =x^2+y^2+{\sin }^2\theta$

Answer 1

Answer: (B)

$x^{2}co{\sec }^2\theta -y^2{\sec }^2\theta =1$

$2A=2\sin \theta$
$A=\sin \theta$
$\Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1(a=2,b=\sqrt{3})\Rightarrow b^2=a^2(1-e^2)$
$3=4(1-e^2)$
$\Rightarrow e^2=1-\frac{3}{4}=\frac{1}{4}\Rightarrow e=\frac{1}{2}$
So $S(ae,0)\Rightarrow S(1,0)$
for hyperbola foci are same
$A{e}_1=ae=1\Rightarrow \left(\sin \theta \right)e_1=1\Rightarrow e_1=co\sec \theta$
$B^2=A^2(e_1^2-1)={\left(A{e}_1\right)}^2-A^2\Rightarrow B^2=1-{\sin }^2\theta ={\cos }^2\theta$
$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1\Rightarrow \frac{x^2}{{\sin }^2\theta }-\frac{y^2}{{\cos }^2\theta }=1\Rightarrow x^{2}co{\sec }^2\theta -y^2{\sec }^2\theta =1$