JEE Advanced 2017 Paper 1

Given: Hyperbola ‌

x2

a2

−‌

y2

16

=1‌‌ . - - - - - - - (1)
Tangent :2x−y+1=0−⋅s−−2 or 2x+1=y‌‌⋅s⋅s⋅s⋅s(2)
As we know, general form of tangent is y=mx+c where condition of tangency ⇒c2= a2m2−b2‌‌......(3)
From Equation of Hyperbola, a=a,b=4
And values of c and m from the given equation of tangent, c=1,m=2
⇒ In Equation (3),⇒12=a2(2)2−(4)2
⇒12=4a2−16
⇒4a2=17⇒a=‌±√17∕2

Now, checking the options with Pythagoras Theorem as they are sides of right angled triangle,


C) 2a,4,1⇒√17,4,1⇒√17>√16⇒√17>4
So, √17 is the bigger side.
By Pythagoras theorem, (1)2+(4)2=(√17)2
1+16=17
It is a correct set.

D) 2a,8,1
8>√17⇒8 is bigger side.
⇒64=(√17)2+(1)2
⇒64≠17+1
So, they cannot be sides of the triangle.

B) a,4,1
4>‌

√17

2

⇒4 is the biggest side.
16=‌

17

4

+1
16≠‌

21

4

They cannot be sides of a right angled triangle.

A) a,4,2
4>‌

√17

2

42=(‌

√17

2

)2+(2)2=‌

17

4

+4
⇒16≠‌

33

4

They cannot be sides of a right angled triangle.