Let C be a curve given by y(x) = 1 + [ 4x - 3 ,x 3 4.] If P is a point on C, such that t…

Question

Let C be a curve given by y(x) = 1 + $$\sqrt {4x - 3} ,x > {3 \over 4}.$$ If P is a point
on C, such that the tangent at P has slope $${2 \over 3}$$, then a point through which the normal at P passes, is :

Accepted Answer

Given,

y = 1 + $$\sqrt {4x - 3} $$

$$ \therefore $$   $${{dy} \over {dx}}$$ = $${1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}$$

$$ \Rightarrow $$   4x - 3 = 9

$$ \Rightarrow $$   x = 3

$$ \therefore $$   y = 1 + $$\sqrt {12 - 3} $$ = 4

$$ \therefore $$   Equation of normal at point P(3,4)

y - 4 = - $${3 \over 2}$$ (x - 3)

$$ \Rightarrow $$   2y - 8 = - 3x + 9

$$ \Rightarrow $$    3x + 2y - 17 = 0

Solutions · 1