Overview
Test Series
In geometry and calculus, one important concept is the normal to a parabola. But what does that really mean?
Let’s break it down. A normal line to a parabola is simply a line that is at a right angle (90°) to the tangent line at a specific point on the parabola.
Now imagine a parabola given by the equation y² = 4ax. Pick any point P on this curve. If you draw a tangent at point P, that line just touches the curve without cutting it. The normal is the line that also passes through point P but goes perpendicular to the tangent.
Get SSC CGL Tier 1 Crash Course (without Books) SuperCoaching @ just
₹4999₹879
Topic | PDF Link |
---|---|
General and Middle Term in Binomial Free Notes PDF | Download PDF |
Circle Study Notes | Download PDF |
Tangents and Normal to Conics | Download PDF |
Increasing and Decreasing Function in Maths | Download PDF |
Wheatstone Bridge Notes | Download PDF |
Alternating Current Notes | Download PDF |
Friction in Physics | Download PDF |
Drift Velocity Notes | Download PDF |
Chemical Equilibrium Notes | Download PDF |
Quantum Number in Chemistry Notes | Download PDF |
Here’s something important to remember:
The product of the slope of the tangent and the slope of the normal is always -1, since they are at right angles.
You can write the equation of a normal line in three main ways:
When you draw a normal line to a parabola, you're drawing a line that is perpendicular to the tangent at a particular point on the curve.
Let’s say you have a parabola and a point P(x₁, y₁) on it. The formula for the normal line at that point depends on the type and direction of the parabola.
Parabola Equation |
Normal at Point (x₁, y₁) |
y² = 4ax |
(y − y₁) = −(y₁ / 2a)(x − x₁) |
y² = −4ax |
(y − y₁) = (y₁ / 2a)(x − x₁) |
x² = 4ay |
(y − y₁) = −(2a / x₁)(x − x₁) |
x² = −4ay |
(y − y₁) = (2a / x₁)(x − x₁) |
Sometimes, instead of using a point on the parabola, we use the slope of the normal line (denoted by m) to write its equation. This is called the slope form of the normal to a parabola.
Let’s take the standard parabola y² = 4ax.
Parabola Equation |
Point of Contact |
Equation of Normal |
y² = 4ax |
(am², –2am) |
y = mx – 2am – am³ |
y² = –4ax |
(–am², 2am) |
y = mx + 2am + am³ |
x² = 4ay |
(–2a/m, a/m²) |
y = mx + 2a + a/m² |
x² = –4ay |
(2a/m, –a/m²) |
y = mx – 2a – a/m² |
When we use a parameter (usually t) to represent a point on a parabola, we can also write the normal to the parabola using this parameter. This form is called the parametric form of the normal.
Let’s start with the standard parabola:
y² = 4ax
Parabola Equation |
Point on Parabola (Parametric Coordinates) |
Equation of Normal |
y² = 4ax |
(at², 2at) |
y = –tx + 2at + at³ |
y² = –4ax |
(–at², 2at) |
y = tx + 2at + at³ |
x² = 4ay |
(2at, at²) |
x = –ty + 2at + at³ |
x² = –4ay |
(2at, –at²) |
x = ty + 2at + at³ |
Let's look at some examples to better understand the concept of the equation of normal to a parabola.
Find the equation of the normal to the parabola y² = 8x at the point (2, 4).
Solution:
The general form of the parabola is:
y² = 4ax
Here, 4a = 8 ⇒ a = 2
For a point (x₁, y₁) on the parabola, the equation of the normal is:
y - y₁ = - (y₁ / 2a)(x - x₁)
Substitute (x₁, y₁) = (2, 4) and a = 2:
y - 4 = - (4 / 4)(x - 2)
y - 4 = - (x - 2)
y - 4 = -x + 2
x + y = 6
Final Answer: x + y = 6
Find the normal to the parabola y² = 12x at the point where y = 6.
Solution:
Parabola is y² = 4ax → Here, 4a = 12 ⇒ a = 3
Let’s find the corresponding x for y = 6:
y² = 12x ⇒ 36 = 12x ⇒ x = 3
Point on the parabola: (3, 6)
Use the normal formula:
y - y₁ = - (y₁ / 2a)(x - x₁)
Substitute values:
y - 6 = - (6 / 6)(x - 3)
y - 6 = - (x - 3)
y - 6 = -x + 3
x + y = 9
Final Answer: x + y = 9
Find the normal to the parabola y² = 4x at the point (1, 2).
Solution:
This is already in standard form with 4a = 4 ⇒ a = 1
Use the normal equation:
y - y₁ = - (y₁ / 2a)(x - x₁)
Substitute (1, 2):
y - 2 = - (2 / 2)(x - 1)
y - 2 = - (x - 1)
y - 2 = -x + 1
x + y = 3
Final Answer: x + y = 3
We hope you found this article regarding Equations of Normal to a Parabola was informative and helpful, and please do not hesitate to contact us for any doubts or queries regarding the same. You can also download the Testbook App, which is absolutely free and start preparing for any government competitive examination by taking the mock tests before the examination to boost your preparation. For better practice, solve the below provided previous year papers and mock tests for each of the given entrance exam:
Download the Testbook APP & Get Pass Pro Max FREE for 7 Days
Download the testbook app and unlock advanced analytics.