Question

For the hyperbola $\frac{(3x-4y-12{)}^2}{100}-\frac{(4x+3y-12{)}^2}{225}=1$

• A. coordinates of center is$(\frac{84}{25},\frac{-12}{25})$
• B. coordinates of center is $(\frac{12}{25},\frac{84}{25})$
• C. coordinates of foci is $(\frac{84\pm 100\sqrt{13}}{25},\frac{-12\mp 75\sqrt{13}}{25})$
• D. coordinates of foci is $(\frac{84\pm 15\sqrt{13}}{25},\frac{-12\mp 20\sqrt{13}}{25})$

Answer 1

Answer: (A), (D)

The correct options are
A coordinates of center is$(\frac{84}{25},\frac{-12}{25})$
D coordinates of foci is $(\frac{84\pm 15\sqrt{13}}{25},\frac{-12\mp 20\sqrt{13}}{25})$

Here equation of the conjugate axis is $3x-4y-12=0$
Equation of the transverse axis is $4x+3y-12=0$
So Centre will be the intersetion point of lines $=(\frac{84}{25},\frac{-12}{25})$
Also $e=\frac{\sqrt{13}}{2}$
Foci are at a distance of $ae=\sqrt{13}$ units from the centre.
So, using parametric form of line, coordinates of foci are $(\frac{84}{25}\pm \sqrt{13}\cos \theta ,\frac{-12}{25}\pm \sqrt{13}\sin \theta )$
where $\tan \theta =\frac{-4}{3}$
$\therefore$ Required foci coordinates are $(\frac{84\pm 15\sqrt{13}}{25},\frac{-12\mp 20\sqrt{13}}{25})$