# RD Sharma Class 12 Chapter 16 Solutions (Tangents And Normals)

RD Sharma Class 12 Maths Solutions for Chapter 16 Tangents and Normals teaches you about the basic and advanced concepts of Tangents and Normals. Experts have designed these RD Sharma Solutions keeping students who are entirely dependent on the web to acquire mathematics of Class 12. The topics covered in this chapter are preliminaries like slope or gradient of a line, slopes of tangent and normal, on finding the slopes of the tangent and the normal at a given point, on finding the point on an available curve where the Tangents are perpendicular or parallel, equations of tangent and normal, algorithm, finding the normal and tangent equation, finding the perpendicular or parallel tangents and normal corresponding to a given line, finding the normal or a tangent that passes through a specific point, angle of intersection of two curves, and orthogonal curves.

Three exercises comprising 49 questions which will help you to get in-depth knowledge of tangents and normals and how we apply them to solve advanced entrance examination questions. In these solutions, you’ll learn different methodologies to apply the concept of tangents and normals in order to solve advanced-level problems.

Instasolv is proud of its experts who provide thorough guidance on higher-level mathematical concepts. Their solutions help you in preparing well for upcoming exams such as CBSE and engineering entrance exams. You’ll also learn the significance of Tangents and Normals in various other chapters of class 12 Mathematics.

## Topics Covered in RD Sharma Solutions for Chapter 16 Tangents and Normals

**Preliminaries**

**The gradient of the Line – **Or slope of the line is formed when that line makes a Trigonometrical tangent with the positive x-axis, in an anti-clockwise direction. You can denote this gradient by m.

Thus, if the above statement is true, then m = tan Θ, where Θ is the angle made by a line with a positive x-axis in an anticlockwise direction.

Since a line falling parallel to x-axis always makes a 0º angle with the x-axis, its slope, tan 0º is equal to 0.

A line that falls perpendicular to the x-axis or a line that falls parallel to y-axis always makes a 90º angle with the axis of x. So the slope tan π/2 = ∞.

Furthermore, a line’s gradient which is equally inclined with axes makes it -1 or +1. This is because it makes either a 45º angle or 135º angle with the axis of x.

Thus the line’s gradient in terms of its coordinates of the two points on the line: Let P(a_{1}, b_{1}) and Q(a_{2}, b_{2}) are the two points on the line. Then its slope c can be calculated by the following formula –

C = (b_{2}-b_{1})/(a_{2}-a_{1}) = (Difference of ordinate)/(Difference of abscissae)

**Calculation of a Slope of a line using the given equation:**

The slope of a line ax + by + c = 0 is

m = -(a/b) = -(Coefficient of x)/(coefficient of y)

The angle between the two lines: Θ is the angle between the two lines of slopes m_{1} and m_{2}.

Then tan Θ = ± ((m_{1}-m_{2})/(1+m_{1}m_{2}))

**Condition of Parallelism: **If the two lines are parallel, then Θ = 0º.

Therefore, tan Θ = tan 0 = 0

=> (m_{2}-m_{1})/(1+m_{1}m_{2}) = 0

=> m_{2} = m_{1}.

This drives us to a conclusion that if the given two lines are parallel, their slopes will also be equal.

**Condition of perpendicularity: **m_{1} and m_{2} are the two slopes of two perpendicular lines, then m_{1}m_{2} will be -1.

m_{1}m_{2} = -1

This means the product of two slopes is -1 if the corresponding two lines are perpendicular.

Furthermore, consider m as a slope of a line, then the slope of the line perpendicular to the first line will be (-1/m).

**Equation of a straight line: **A straight line passing through a point (a_{1},b_{1}) with a slope m is

(b – b_{1}) = m(a – a_{1})

### Discussion of exercises in RD Sharma Class 12 Maths Solutions Chapter 16 Tangents and Normals

- Exercise 16.1 checks your knowledge on tactics and methods you use in finding the slope of tangents and normals for a given curve. The exercise also asks about finding the values of two variables of a given curve. You will also come across solutions that evaluate points on a given curve when the tangent is parallel or perpendicular to a given line.
- Exercise 16.2 talks about finding the equation of a normal or a tangent to a given curve.
- Exercise 16.3 tests your knowledge by asking you questions such as finding the intersection angle or proving that a given set of curves are intersecting orthogonally and many more.

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