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=0 gives =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,) Perpendicular distance between line
x−y=1 and point
(1,) is given as,
P=|| (1)(1)+()(−1)−1 |
| √(1)2+(−1)2 |
| =|| =∴ using (
∴ using perpendicular distance formula)