JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q73

← JEE Main 2016 · 10 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 73 of 90

Let C be a curve given by $y(x)=1+\sqrt{4x-3}$ , $x>\dfrac34$ . If P is a point on C, such that the tangent at P has slope $\dfrac23$ , then a point through which the normal at P passes, is :

(2, 3)

(4, −3)

(1, 7)

(3, −4)

Answer and worked solution

Official answer: C

$y'=\frac2{\sqrt{4x-3}}=\frac23$ gives x = 3, y = 4. The normal $y-4=-\frac32(x-3)$ passes through (1, 7).

Topics:Tangents and normals

← Previous question Next question →

JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 73