CGPET 2022 Solved Paper

We have,
4x2−9y2=144
⇒

x2

36

−

y2

16

=1
which is equation of hyperbola.
Now, equation of asymptotes of hyperbola

x2

a2

−

y2

b2

=1 is y=±‌

b

a

x
∴ Equation of asymptotes of given hyperbola are
y=±

4

6

x
⇒y=±

2

3

x
So, equation of asymptotes are 3y+2x=0 and 3y−2x=0
Let (h,k) be any point of hyperbola. Then,
Distance of (h,k) from 3y+2x=0 is given by
d1=|

2h+3k

√4+9

|=|

2h+3k

√13

|
and distance of (h,k) from 3y−2x=0 is given by
d1=|

2h+3k

√4+9

|=|

2h+3k

√13

| and distance of (h,k) from 3y−2x=0 is given by
d2=|

−2h+3k

√4+9

|
=|

−2h+3k

√13

|
Now, d1d2=|

2h+3k

√13

|×|

−2h+3k

√13

|
=|

−4h2+9k2

13

|
=|

−(4h2−9k2)

13

| ...[∵(h,k) lies on hyperbola]
=|

−144

13

|=‌

144

13