AIEEE 2012 Solved Paper

Given equation of ellipse is 2x2+y2=4
⇒‌

2x2

4

+‌

y2

4

=1⇒‌

x2

2

+‌

y2

4

=1
Equation of tangent to the ellipse ‌

x2

2

+‌

y2

4

=1 is
y=mx±√2m2+4.......(1)
( as equation of tangent to the ellipse ‌

x2

a2

+‌

y2

b2

=1
is y=mx+c where c=±√a2m2+b2 )
Now, Equation of tangent to the parabola
y2=16√3x is y=mx+‌

4√3

m

......(2)
( as equation of tangent to the parabola
y2=4ax‌ is ‌y=mx+‌

a

m

‌ ) ‌
On comparing (1) and (2), we get
‌

4√3

m

=±√2m2+4
Squaring on both the sides, we get
‌16(3)=(2m2+4)m2
‌⇒48=m2(2m2+4)⇒2m4+4m2−48=0
‌⇒m4+2m2−24=0⇒(m2+6)(m2−4)=0
‌⇒m2=4‌ ( as ‌m2≠−6)⇒m=±2
⇒ Equation of common tangents are y=±2x±2√3
Thus, statement - 1 is true.
Statement - 2 is obviously true.