EAMCET Engineering 2010 Solved Paper

Given, curve is
2x2+y2=2x
2x2−2x+y2=0
⇒‌‌2(x−‌

1

2

)2+y2=‌

1

2

⇒‌‌‌

(x− 1 2 )2

1 4

+‌

y2

1 2

=1
which represents an ellipse.
Here, a=‌

1

2

,b=‌

1

√2

,h=‌

1

2

,k=0
Consider a point P(h+a‌cos‌θ,k+bsin‌θ) =P(‌

1

2

+‌

1

2

‌cos‌θ,‌

1

√2

sin‌θ) on the ellipse from which the distance of point (a,0) is maximum. let‌Q(a,0)
Now,
‌PQ=√‌

1

2

+a2−a+(‌

1

2

−a)‌cos‌θ+‌

1

4

sin‌2θ
‌⇒‌ Let ‌y=PQ2
‌=‌

1

2

+a2−a+(‌

1

2

−a)‌cos‌θ+‌

1

4

sin‌2θ
For maxima and minima, put ‌

dy

dθ

=0
‌⇒−(‌

1

2

−a)sin‌θ+‌

1

4

⋅2sin‌θ‌cos‌θ=0
‌⇒‌‌sin‌θ(−‌

1

2

+a+‌

1

2

‌cos‌θ)=0
‌⇒sin‌θ=0‌ or ‌−‌

1

2

+a+‌

1

2

‌cos‌θ=0
‌⇒‌‌θ=0‌ or ‌cos‌θ=1−2a
‌⇒‌‌sin‌2θ=1−cos2θ
‌=1−(1−2a)2
‌=−4a2+4a
Now, ‌

d2y

dθ2

<0 for cos‌θ=1−2a
Thus, distance PQ is maximum, when cos‌θ=1−2a and sin‌2θ=−4a2+4a
Now, required longest distance is
=‌√‌

1

2

+a2−a+(‌

1

2

−a)(1−2a)+‌

1

4

(−4a2+4a)
‌=√‌

1

2

+a2−a+‌

1

2

−a−a+2a2−a2+a
‌=√2a2−2a+1
‌=√1−2a+2a2