NIMCET 2018 Question Paper with solutions

Let the equation of line y - mx + c ... (i)
If Eq. (i) is tangent to the hyperbola x29−

y2

4

= 1
∴ c = ± √9m2−4 [a = 3 , b = 2]
So, equation of line (i) is
y = mx ± √9m2− ... (ii)
It is also tangent to the circle x2+y2−8x=0 here, centre
c - (4, 0) and radius (r) = 4
Perpendicular distance from centre to the tangent = Radius
∴

|4m±√9m2−4|

√1+m2

= 4
(4m±√9m2−)2 = 16 (1+m2) ... (iii) [squaring both sides]
By solving Eq. (iii), we get
m =

2

√5

Put the value of m =

2

√5

in eq . (i)
∴ y =

2

√5

x± √9×

4

5

- 4
=

2

√5

x±

√16

5

=

2x

√5

±

4

√5

⇒ √5y = 2x ± 4
⇒ 2x ± 4 - √5y = 0
⇒ 2x - v5y + 4 = 0