Let C be a curve given by yx - 1 - -4x - 3- x - 34- If P is a point on C- such that the tangent at P has slope 23- then a point through which the normal at P passes- is-

The correct option is D (1, 7) dydx = 12√(4x 3)× 4 = 23 ⇒ 4x 3 = 9 ⇒ x = 3 So, y = 4 Equation of normal at P(3, 4) is y ...

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Standard XII

Mathematics

Equation of Tangent at a Point (x,y) in Terms of f'(x)

Let C be a ...

Question

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:

(1, 7)

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(3, - 4)

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(4, - 3)

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(2, 3)

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Solution

The correct option is D $(1, 7)$
$\frac{d y}{d x} = \frac{1}{2 \sqrt{4 x - 3}} \times 4 = \frac{2}{3}$
$\Rightarrow 4 x - 3 = 9$
$\Rightarrow x = 3$
So, $y = 4$
Equation of normal at $P (3, 4)$ is
$y - 4 = - \frac{3}{2} (x - 3)$
i.e. $2 y - 8 = - 3 x + 9$
$\Rightarrow 3 x + 2 y - 17 = 0$

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Let C be a curve given by y(x)=1+√4x−3,x>34. If P is a point on C, such that the tangent at P has slope 23, then a point through which the normal at P passes, is:

The correct option is D (1,7)dydx=12√4x−3×4=23⇒4x−3=9⇒x=3So, y=4Equation of normal at P(3,4) isy−4=−32(x−3)i.e. 2y−8=−3x+9⇒3x+2y−17=0

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:

The correct option is D $(1, 7)$
$\frac{d y}{d x} = \frac{1}{2 \sqrt{4 x - 3}} \times 4 = \frac{2}{3}$
$\Rightarrow 4 x - 3 = 9$
$\Rightarrow x = 3$
So, $y = 4$
Equation of normal at $P (3, 4)$ is
$y - 4 = - \frac{3}{2} (x - 3)$
i.e. $2 y - 8 = - 3 x + 9$
$\Rightarrow 3 x + 2 y - 17 = 0$