AMU Engineering 2018 solved paper Solved Paper

We have,
2y2=2x−1 and 2x2=2y−1
⇒y2=x−

1

2

and x2=y−

1

2

Now, the shortest distance always along common normal to curve.
Equation of normal to the curve y2=x−

1

2

is
y=m(x−

1

2

)−2×

1

4

m−

1

4

m3
⇒y=mx−

1

2

m−

1

2

m−

1

4

m3
⇒y=mx−m−

m3

4

........(i)
Equation of normal to the curve x2=y−

1

2

is
y−

1

2

=mx+2(

1

4

)+

1

4m2

⇒y=mx+1+

1

4m2

.........(ii)
Since, Eqs. (i) and (ii) are same normal.
∴−m−

m3

4

=1+

1

4m2

⇒

−4m−m3

4

=

4m2+1

4m2

⇒m5+4m3+4m2+1=0
⇒(m+1)(m4−m3+5m2−m+1)=0
⇒m=−1
Hence, slope of tangent =1
Equation of tangent to curve y2=x−

1

2

is
y=(x−

1

2

)+

1

4

⇒y=x−

1

4

Equation of tangent to curve x2=y−

1

2

is
y=x+

1

4

∴ Required distance =|

1 4 + 1 4

√2

|=

1

2√2