AP EAMCET Engineering 2012 Solved Paper

Given hyperbola is 5x2−y2=5
or It can be rewritten as

x2

1

−

y2

5

=1
Here, a2=1,b2=5
∴ Equation of tangent is
y=mx±√a2m2−b2
⇒ y=mx±√1m2−5 ......(i)
But point (2,8) lies on it.
∴8=2m±√m2−5
⇒(8−2m)=±√m2−5
On squaring both sides, we get
64+4m2−32m=m2−5
⇒ 3m2−32m+69=0
⇒ (3m−23)(m−3)=0
⇒ m=3,

23

3

On putting m=3 in Eq. (i), we get
y&=3x±√32−5=3x±2
⇒y=3x+2and y=3x−2