Manipal Entrance Test 2015 Math Paper

At the point where the given curve crosses X-axis,
we have
y=0
⇒ ax2=1
[putting y=0 in ax2+2hyx+by2=1 ]
⇒ x=±

1

√a

Thus, the given curve cuts X -axis at P

1

√a

,0) and Q (−

1

√a

,0 )
Now, ax2+2hxy+by2=1
On differentiating both sides w.r.t. x, we get
2ax+2h(x

dy

dx

+y )+2by

dy

dx

=0
⇒

dy

dx

=−

ax+hy

hx+by

⇒ [

dy

dx

]P=−

a

h

and [

dy

dx

]a=−

a

h

⇒ [

dy

dx

]P=[

dy

dx

]a
Hence, the tangents are parallel.