Let (x,y) be any arbitrary point on curve x2=2y and find the tangent line equation at this point, such that tangent line at (x,y) is parallel to line x−y=1. To find tangent equation, differentiate the following equation so that we can find slope, x2−2y‌‌=0 2x−2‌
dy
dx
‌‌=0‌ gives ‌‌
dy
dx
=x Slope ( say m1)=x Also, slope of line x−y=1 or y=x−1 is 1 (say m2 ). Since, x−y=1 and tangent line is parallel, therefore, their slope be equal. Hence, m1=m2 gives, x=1 Put x=1 in Eq. (i), we get y=1∕2 Thus, (x,y)=(1,‌
1
2
) Perpendicular distance between line x−y=1 and point (1,‌
1
2
) is given as, P‌‌=|‌
(1)(1)+(
1
2
)(−1)−1
√(1)2+(−1)2
| ‌‌=|‌
−1
2√2
| ‌‌=‌
1
2√2
‌‌‌‌‌‌∴‌ using ‌ ( ∴ using perpendicular distance formula)