The vertices of a hyperbola are at (0,0) and (10,0) and one of its foci is at (18,0). The possible equation of the hyperbola is -

Question

The vertices of a hyperbola are at (0,0) and (10,0) and one of its foci is at (18,0). The possible equation of the hyperbola is - (A) (x2/25)-(y2/144)=1 (B) ((x-5)2/25)-(y2/144)=1 (C) (x2/25)-((y-5)2/144)=1 (D) ((x-5)2/25)-((y-5)2/144)=1

Tardigrade Admin · Accepted Answer

Centre of hyperbola is (5,0), so equation is ((x-5)2/a2)-(y2/b2)=1 a=5, a e-a=8 ⇒ e=(13/5) b2=144 So equation is ((x-5)2/25)-(y2/144)=1.

Q. The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$ . The possible equation of the hyperbola is -

$\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1$

$\frac{( x - 5 ) ^{2}}{25} - \frac{y ^{2}}{144} = 1$

$\frac{x ^{2}}{25} - \frac{( y - 5 ) ^{2}}{144} = 1$

$\frac{( x - 5 ) ^{2}}{25} - \frac{( y - 5 ) ^{2}}{144} = 1$

Centre of hyperbola is $(5, 0)$ , so equation is
$\frac{( x - 5 ) ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
$a = 5, a e - a = 8 \Rightarrow e = \frac{13}{5}$
$b^{2} = 144$
So equation is $\frac{( x - 5 ) ^{2}}{25} - \frac{y ^{2}}{144} = 1$ .

Q. The vertices of a hyperbola are at (0,0) and (10,0) and one of its foci is at (18,0). The possible equation of the hyperbola is -

25x2​−144y2​=1

25(x−5)2​−144y2​=1

25x2​−144(y−5)2​=1

25(x−5)2​−144(y−5)2​=1