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

Question

Let C be a curve given by y(x) = 1 + √4x - 3 , x > (3/4). If P is a point on C, such that the tangent at P has slope (2/3), then a point through which the normal at P passes, is : (A) (2, 3) (B) (4, -3) (C) (1, 7) (D) (3, -4)

Tardigrade Admin · Accepted Answer

y(x)=1+√4 x-3, x>(3/4) Let P (α, 1+(√4 α-3) be the point. at which (d y/d x) text ATP =(2/3) ⇒ (2/√4 α-3)=(2/3) ⇒ 4 α-3=9 ⇒ α=3 Hence P (3,4) slope of normal at P(3,4) is =-(3/2) equation of normal Y-4=-(3/2)(X-3) 2 y-8=-3 x+9 3 x+2 y=17 clearly it is passes through (1,7)

Q. Let C be a curve given by y(x)=1+4x−3​,x>43​. If P is a point on C, such that the tangent at P has slope 32​, then a point through which the normal at P passes, is :

(2, 3)

(4, -3)

(1, 7)

(3, -4)

y(x)=1+4x−3​,x>43​ Let P(α,1+(4α−3​) be the point. at which dxdy​ATP ​=32​ ⇒4α−3​2​=32​ ⇒4α−3=9 ⇒α=3 Hence P(3,4) slope of normal at P(3,4) is =−23​ equation of normal Y−4=−23​(X−3) 2y−8=−3x+9 3x+2y=17 clearly it is passes through (1,7)

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