AP EAPCET 18 May 2024 Shift 1 Solved Paper

We have, equarion of hyperbola
‌5x2−ky2=12
‌‌

x2

12 5 −‌ y2 12

=1
Here, a2=‌

12

5

and b2=‌

12

K

Equation of tangent of hyperbola is
5x−2y−6=0
or y=‌

5

2

x−3
Here, m=‌

5

2

and a2m2−b2=9
‌

12

5

×‌

25

4

−‌

12

K

=9⇒15−9=‌

12

K

K=2
Equation of hyperbola is
‌5x2−2y2=12
‌‌

x2

12 5

−‌

y2

6

=1
Since, (√6,p) lie on hyperbola
‌⇒5(√6)2−2(p3)=12
‌⇒‌‌30−2p2=12
‌‌‌p2=9‌‌p=23
‌∵‌‌p<0
‌∴‌‌p=−3
Equation of normal of hyperbola of (√6,−3),3
‌⇒‌‌‌

12x

5√6

−2y=‌

12

5

+6
‌⇒‌‌2√6x−10y=42
‌⇒‌‌√6x−5y=42