BITSAT 2021 Paper

y2=x(2−x)2⇒y2=x3−4x2+4x ...(i)
⇒ 2y‌

dy

dx

=3x2−8x+4
⇒

dy

dx

=

3x2−8x+4

2y

⇒ [

dy

dx

]P=

3−8+4

2

=−

1

2

∴ Equation of tangent at P is: y−1=−

1

2

(x−1)
⇒ x+2y−3=0
Using y=

3−x

2

in (i), we get: (

3−x

2

)2
=x3−4x2+4x
⇒ 4x3−17x2+22x−9=0 ..(ii)
which has two roots 1,1(Because of (ii) being tangent at (1,1) ).
Sum of 3 roots =

17

4

∴3‌rd root =

17

4

−2=

9

4

Then, y=

3− 9 4

2

=

3

8

∴ Q is (

9

4

,

3

8

)