Question

OR
Find the point on the curve y=
$\sqrt{x - 3}$
, where the tangent is perpendicular to the line
6x + 3y - 5 == 0​

Answer 1

Step-by-step explanation:

Given curve y=√x-3

Given line 6x + 3y - 5 = 0

=>3y = 5 - 6x

=>y = 5-6x/3

=> - 6/3 x + 5/3

m2 = -6/3

= -2

m1 × (-2) = -1

=> m1=1/2

y = √x-3

dy/dx = 1/2 √x - 3

=> 1/2√x - 3 = 1/2

=> √x - 3 = 1

=> x - 3 = 1

=> x = 4

y = √4 - 3 = 1

= point on the curve is (4,3)