Students who dream of pursuing their engineering in IITs and other premier engineering colleges in India should clear the JEE Main and the JEE Advanced. Since a huge number of candidates appear for the exam, the level of competition is very high. One needs complete dedication and hard work to crack the JEE exam. Students should plan and prepare accordingly so that they can secure an IIT seat.
As far as the JEE Main exam is concerned, differentiation and application of derivatives are topics of great importance. Tangents and normals come under the topic of application of derivatives. Students can definitely expect one question from this topic. So they are recommended to have thorough knowledge about this topic. In this article, we will discuss the tangent, normal, equation of tangent and equation of normal.
Tangent and Normal
A tangent to a curve is a line that touches the curve at just one point. The normal to the curve is the line perpendicular to the tangent to the curve at that point. The product of the slope of normal and slope of the tangent is equal to -1. The equation of a line passing through the point (x1, y1) having slope m is given by y-y1 = m(x-x1).
Slope of Tangent
Let y = f(x) be a curve. We have to find the slope of the tangent to the curve at the point (x1, y1). First, we differentiate the equation of the curve y = f(x), i.e. we find dy/dx. Then put (x1, y1) in dy/dx. So dy/dx at (x1, y1) gives the slope of the tangent.
Equation of normal
We know that tangent is perpendicular to the normal. If m1 is the slope of the normal, and m2 is the slope of the tangent, then the product m1m2 = -1.
Therefore, the slope of the normal to the curve y = f(x) at the point (x1, y1) is given by -1/(dy/dx) at (x1, y1), [provided (dy/dx) at (x1, y1) is not equal to zero].
So, the equation of the normal to the curve y = f(x) at the point (x0, y0) is
y-y1 = (-1/(dy/dx)(x1, y1) )(x – x1).
Let us discuss an example of finding the equation of tangent and normal.
Example: Given curve y = x3. Find the equations of the tangent and normal to the given curve at the point (1, 1).
Given y = x3.
Differentiating we get
dy/dx = 3x2
(dy/dx) at (1, 1) = 3.
Slope of tangent = 3
Equation of tangent is y – 1 = 3(x – 1)
=> y – 1 = 3x – 3
=> y = 3x – 2
Slope of normal = -1/3
The equation of normal is y – 1 = (-⅓)(x – 1).
=> -3y + 3 = x – 1
=> -3y – x = -1 – 3
=> -3y – x = -4
=> x + 3y = 4
A curve obtained when a plane intersects a double right circular hollow cone is termed a conic section. Parabola, hyperbola, ellipse, and circle are the important conic sections which students should learn for the JEE exam. A conic section is an important topic for any engineering entrance exam. If the students are thorough with the basics of the conic section, then they can easily crack problems from the conic section.
Visit BYJU’S to learn more about the conic section, important formula PDFs of the conic section, previous years’ solved JEE question papers and much more.