Let C be a curve given by yx=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 :

Correct answer: (A) 1, 7. Given, yx=1+ √(4x-3) So, dy/dx=1/(2√(4x-3)) ×4=2/3 ⇒ 4x-3=9 ⇒ x=3 So, y=4 Equation of normal at P3, 4 is y-4=-3/2x-3 i.e., 2y-8=-3x+9 ⇒ 3x+2y-17=0 Hence only option A is satisfying the equation.

Q55 / 90 · JEE Main 2016 (10 Apr Online) · Mathematics

Let C be a curve given by $y (x) = 1 + \sqrt{4 x - 3}$ , $x > \frac{3}{4}$ . 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 :

Correct answer
A
$(1, 7)$
B
$(3, - 4)$
C
$(4, - 3)$
D
$(2, 3)$

Solution

Given,

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

So, $\frac{d y}{d x} = \frac{1}{2 \sqrt{4 x - 3}} \times 4 = \frac{2}{3}$

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

$\Rightarrow$ $x = 3$

So, $y = 4$

Equation of normal at $P (3, 4)$ is $y - 4 = - \frac{3}{2} (x - 3)$

i.e., $2 y - 8 = - 3 x + 9$

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

Hence only option $A$ is satisfying the equation.

Hence, the correct answer is option: 'A'.

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Solution

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Let C be a curve given by $y (x) = 1 + \sqrt{4 x - 3}$ , $x > \frac{3}{4}$ . 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 :