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-

Y = 1 + √(4x – 3) dydx=12√(4x 3)× 4=23 ⇒ x=3, y=1+3=4 Slope of the normal= 32 Equation of the normal is, y 4= 32(x 3) ⇒ 3x+2y 17=0 ⇒Which passes through (1, ...

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Standard XII

Mathematics

Solving Linear Differential Equations of First Order

Let C be a cu...

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:

(3, - 4)

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(1, 7)

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(2, 3)

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(4, - 3)

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Solution

The correct option is B $(1, 7)$
$y = 1 + \sqrt{4 x - 3} \frac{d y}{d x} = \frac{1}{2 \sqrt{4 x - 3}} \times 4 = \frac{2}{3} \Rightarrow x = 3, y = 1 + 3 = 4 Slope of the normal = - \frac{3}{2} Equation of the normal is, y - 4 = - \frac{3}{2} (x - 3) \Rightarrow 3 x + 2 y - 17 = 0 \Rightarrow Which passes through (1, 7)$

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Solving Linear Differential Equations of First Order

Standard XII Mathematics

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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 B (1,7) y=1+√4x–3dydx=12√4x−3×4=23⇒x=3,y=1+3=4Slope of the normal=−32Equation of the normal is,y−4=−32(x−3)⇒3x+2y−17=0⇒Which passes through (1,7)

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: