AP EAPCET 05-Jul-2022 Shift 1 Paper

Let (x,y) be on the parabola y=x2+4x+3 which is closet to the line y=3x+2
Perpendicular distance between a point (x1,y1) and a line (ax+by+c=0)
D=|‌

ax1+by1+c

√a2+b2

|
[∴. line 3x−y+2=0 at point (x1,y1)]
⇒D‌=‌

|−x2−x−1|

√10

⇒‌‌D‌=‌

x2+x+1

√10

=‌

(x+ 1 2 )2+ 3 4

√10

On differentiating w.r.t. x,
‌‌

dD

dx

=‌

(x+ 1 2 )2

√10

+‌

‌ 3 4

√10

⇒‌

dD

dx

=0,x=‌

−1

2

⇒y=x2+4x+3
⇒y=‌

1+12−8

4

⇒y=‌

5

4

‌‌

dD

dx

=‌

2(x+ 1 2 )

√10

⇒‌

d2D

dx2

>0
Hence, minimum at (‌

−1

2

,‌

5

4

).