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BITSAT 2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 110 of 130

Marks: +1, -0

The vertices of the hyperbola

9 x^2-16 y^2-36 x+96 y-252=0

$(6,3),(-2,3)$
$(6,3),(-6,3)$
$(-6,3),(-6,-3)$
$(2,3),(-2,3)$

Solution: 👈: Video Solution

Given hyperbola is

\; 9 x^2-16 y^2-36 x+96 y-252=0

\; \Rightarrow 9 (x^2-4 x) -16 (y^2-6 y) =252

\; \Rightarrow 9 (x^2-4 x+4) -16 (y^2-6 y+9)

\; =252+36-144

\; \Rightarrow 9(x-2)^2-16(y-3)^2=144

\; \Rightarrow \; \frac{(x-2)^2}{16} - \; \frac{(y-3)^2}{9}=1

. . . (i)

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)

.

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