Question

For the hyperbola $3y^2-16x^2-12y+32x-52=0$ the correct option(s) is/are

• A. its length of transverse axis is $8$ units
• B. its length of conjugate axis is $2\sqrt{3}$ units
• C. end points of one latus rectum will be $(\frac{1}{4},2+\sqrt{19})$ and $(\frac{7}{4},2+\sqrt{19})$
• D. Distance between the two directrices is $\frac{16}{\sqrt{19}}$ units

Answer 1

Answer: (A), (B), (C)

end points of one latus rectum will be $(\frac{1}{4},2+\sqrt{19})$ and $(\frac{7}{4},2+\sqrt{19})$

$3y^2-16x^2-12y+32x-52=0$
$\Rightarrow 3(y-2{)}^2-16(x-1{)}^2=48$
$\Rightarrow \frac{(y-2{)}^2}{16}-\frac{(x-1{)}^2}{3}=1...\left(1\right)$
This equation represents a hyperbola whose centre is at $(1,2)$
and eccentricity
$e=\sqrt{1+\frac{(\sqrt{3}{)}^2}{4^2}}=\frac{\sqrt{19}}{4}$
Here,
$b^2=16\Rightarrow b=4,$
$a^2=3,\Rightarrow a=\sqrt{3}$.
$\therefore$ length of transverse axis $=2b=8$ units
and length of conjugate axis $=2a=2\sqrt{3}$ units.
Distance between the two directrices $=\frac{2b}{e}=\frac{32}{\sqrt{19}}$ units

Coordinates of foci will be $(1,2\pm be)=(1,2\pm \sqrt{19})$
End points of latus rectum will be
$y-$ coordinate of latus rectum will be $2\pm \sqrt{19}$
Putting value of $y$ in eqn $\left(1\right)$, we get
$x=\frac{1}{4},\frac{7}{4}$
end points of one latus rectum will be $(\frac{1}{4},2+\sqrt{19})$, $(\frac{7}{4},2+\sqrt{19})$, $(\frac{1}{4},2-\sqrt{19})$ and $(\frac{7}{4},2-\sqrt{19})$