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CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q22: Normal to y=x+1xy = x + \frac1x perpendicular to 3x−4y=73x - 4y = 7

Sample 2020-21Ch 6 · Application of Derivatives
The question

Find the equation of the normal to the curve y=x+1xy = x + \dfrac{1}{x}⁠, x>0x > 0 perpendicular to the line 3x−4y=73x - 4y = 7⁠.

This question uses the normal to a curve. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer
y=x+1x⇒dydx=1−1x2y = x + \frac1x \Rightarrow \frac{dy}{dx} = 1 - \frac{1}{x^2}
Why?

Write 1x=x−1\frac1x = x^{-1} and use the power rule:

ddx(x−1)=−x−2=−1x2\frac{d}{dx}\left(x^{-1}\right) = -x^{-2} = -\frac{1}{x^2}

dydx\frac{dy}{dx} at a point is the slope of the tangent there.

The normal is perpendicular to 3x−4y=73x - 4y = 7⁠, so the tangent is parallel to it. The slope of the line is 34\frac34⁠.

Why?

The tangent is at right angles to the normal, and so is the given line. Two lines both perpendicular to the normal are parallel, so they have the same slope.

Making yy the subject gives the slope of the line:

3x−4y=7−4y=−3x+7y=−3x−4+7−4y=34x−74\begin{aligned} 3x - 4y &= 7 \\ -4y &= -3x + 7 \\ y &= \frac{-3x}{-4} + \frac{7}{-4} \\ y &= \frac34x - \frac74 \end{aligned}

The slope is the number in front of xx⁠: 34\frac34⁠.

1−1x2=34⇒x2=41 - \frac{1}{x^2} = \frac34 \Rightarrow x^2 = 4
Why?

Set the slope of the tangent equal to 34\frac34⁠. Then 1x2=1−34=14\frac{1}{x^2} = 1 - \frac34 = \frac14⁠, so x2=4x^2 = 4 and x=±2x = \pm 2⁠.

1−1x2=341x2=1−34=44−34=14x2=4(flip both sides)x=±4=±2\begin{aligned} 1 - \frac{1}{x^2} &= \frac34 \\ \frac{1}{x^2} &= 1 - \frac34 \\ &= \frac44 - \frac34 = \frac14 \\ x^2 &= 4 \quad (\text{flip both sides}) \\ x &= \pm\sqrt4 = \pm 2 \end{aligned}

Watch out: the curve is given only for x>0x > 0⁠, so x=−2x = -2 is rejected.

⇒x=2\Rightarrow x = 2 (∵x>0)(\because x > 0)

When x=2x = 2⁠, y=2+12=52y = 2 + \frac12 = \frac52⁠.

Why?

Put x=2x = 2 into the curve y=x+1xy = x + \frac1x⁠:

y=2+12=42+12=52\begin{aligned} y &= 2 + \frac12 \\ &= \frac42 + \frac12 = \frac52 \end{aligned}

The slope of the normal is −43-\frac43⁠, so the equation of the normal is

Why?

The normal is perpendicular to the tangent, and the slopes of perpendicular lines multiply to −1-1⁠:

m=−1(34)=−1×43=−43\begin{aligned} m &= -\frac{1}{\left(\frac34\right)} \\ &= -1 \times \frac43 = -\frac43 \end{aligned}

Then use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (2,52)\left(2, \frac52\right)⁠.

y−52=−43(x−2)⇒6y−15=−8x+16\begin{aligned} y - \frac52 &= -\frac43(x - 2) \\ \Rightarrow 6y - 15 &= -8x + 16 \end{aligned}
Why?

Multiplying both sides by 6 clears the fractions: 6×52=156 \times \frac52 = 15 and 6×43=86 \times \frac43 = 8⁠. Then move −8x-8x to the left and −15-15 to the right: 16+15=3116 + 15 = 31⁠.

6y−6(52)=−6(43)(x−2)6y−15=−8(x−2)6y−15=−8x+168x+6y−15=168x+6y=16+15=31\begin{aligned} 6y - 6\left(\tfrac52\right) &= -6\left(\tfrac43\right)(x - 2) \\ 6y - 15 &= -8(x - 2) \\ 6y - 15 &= -8x + 16 \\ 8x + 6y - 15 &= 16 \\ 8x + 6y &= 16 + 15 = 31 \end{aligned}

∴\therefore Equation of the normal: 8x+6y=318x + 6y = 31

Where this comes from

Chapter 6, Application of Derivatives: Normal to a curve perpendicular to a given line.

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