JEE Advanced 2020 Full Test 5 Paper 1

Curve is (y−2)2=2x3−4
‌

dy

dx

=‌

3x2

y−2

∴ Equation of tangent at (x1,y1)
y−y1=‌

3x12

y1−2

(x−x1)
As it passes through (1,2)
⇒‌‌2−y1=‌

3x12

y1−2

(1−x1)
‌ or ‌‌‌(y1−2)2=3x12(x1−1)
or 2x13−4=3x13−3x12[ As (x1,y1) lies on the curve ]
or x13−3x12+4=0
or (x1+1)(x1−2)2=0
∴‌‌x1=−1 or 2
But x1≠=−1 (it is clear from equation (1))
∴x1=2
∴y1=2±2√3
Points of contact are (2,2+2√3) and (2,2−2√3)
∴ sum of x -coordinates of points of contact is 4
∴ sum of y -coordinates of points of contact is 4