Exam

BITSAT

Year

2023

Paper

Paper 2

Subject

Mathematics

Chapter

Hyperbola

BITSAT 2023 · Paper 2 · Mathematics · Hyperbola

The vertices of the hyperbola 9 x^2-16 y^2-36 x+96 y-252=0 are

The vertices of the hyperbola $9 x^2-16 y^2-36 x+96 y-252=0$ are

Step-by-step solution

Given hyperbola is
$$\begin{aligned} \\ & 9 x^2-16 y^2-36 x+96 y-252=0 \\ \\ & \Rightarrow 9\left(x^2-4 x\right)-16\left(y^2-6 y\right)=252 \\ \\ & \Rightarrow 9\left(x^2-4 x+4\right)-16\left(y^2-6 y+9\right) \\ \\ & =252+36-144 \\ \\ & \Rightarrow 9(x-2)^2-16(y-3)^2=144 \\ \end{aligned}$$


Put $x=X+2 and y=Y+3$
$\therefore$ Equation (i) becomes $\frac{X^2}{16}-\frac{Y^2}{9}=1$
Now, vertices are $X= \pm a$ wehre $a=4 and Y=0$
Hence, vertices are $(6,3),(-2,3)$.