JEE advanced important topics : Tangent & Normal Of the Parabola
JEE advanced important topics : The parabola is an open and flat curve, which is defined as the geometric place of the points of the plain that are equidistant of a point named focus (F), and a line named directive (D), from the figure we can see that, FP = PQ = r.
The axis of the parabola is the perpendicular line of the directive, that goes through focal point F. The distance FD, of the focal to the directive, it’s called parameter of the parabola and the point in the middle of the segment FD is the V point, which is named vertex of the parabola.
Canonical or reduced equation of the parabola
Let us consider a parabola which axis is the y axis, it’s vertex V is in (0 , 0) and it’s in the in the positive x space.
In this case the focus will necessary be in F (p/2,0). The equation of the directive line D will be x=-p/2.
The radial vectors FP and PM that correspond to any point P of the parabola (and are, by definition, the same) will have the longitude:
Simply put, we obtain the canonical equation or reduced equation of the parabola, referred to this configuration:
If one moves the E axis parallel to the y axis and the vertex of the parabola to the point V = (xV,yV), the equation of this parabola is the same as the one shown on the image, which also shows the formula for the directive line D.
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Canonical equation but now with it’s axis coinciding with the x axis, it’s vertex is the center of coordinates V (0,0) and it is in the positive y space.
If you move the E axis parallel to the x axis and the vertex of the parabola to the point V(xV,yV), the equation of this parabola will be the same as the one shown on the image.
In the next example we will study the same canonical or reduced equation for a parabola whose axis coincide with either the y or the x axis, always with the vertex in the origin O (0,0), but now with negative values for x an y respectively.
The formula for the parabola with V vertex in V (xV, yV) and vertical axis parallel to the OY axis is:
As long as a =/=0 and b and c are real numbers, the value of the constants is:
The constant a shows how “open” the parabola is. When the value of a decreases, the parabola opens. You could also say that the parabola will be more open when the p value increases. If the a constant is positive, the vertex V will be the lowest point in the parabola, which means it will be open upwards, like bowl of soup. If the constant a is negative, the vertex V will be the peak of the parabola, which means the parabola will open downwards, like an igloo.
The equation of the parabola with vertex V (xV,yV) and the horizontal axis parallel to the OX axis is:
In which it’s constants equal to:
The constant a shows how “open” the parabola is. When the value of a decreases, the parabola opens. You could also say that the parabola will be more open when the p value increases. If the a constant is positive, it will be open to the right, like the letter “C”. If the constant a is negative, the parabola will open to the left.
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General formula of the parabola
We have studied parabolas that have vertical or horizontal axis, and are particular cases of the general formula of the parabola:
As long as
– AB = 0 and A & B are not equal to zero at the same time.
Let us study more instances of parabolas in which it’s axis are neither horizontal nor vertical. In the case of the inclined parabola:
In this four cases it’s equation has all the terms of the general equation of the parabola.
We can corroborate that in the four parabola – AB = 0 and A & B are not equal to zero at the same time.
JEE advanced important topics : Normal to the Parabola
The normal to the parabola, also called slope of the tangent line to the parabola, at a given point is the derivative of the function of the parabola at said point.
Tangent line of the parabola in a certain point
The tangent line to a parabola in a given point is the one that cuts though the point (a, f(a)) and which slope is equal to f ‘(a). In other words, it’s formula is given by:
Find the equation of the tangent line to the parabola y = x² − 5x + 6, parallel to the line 3x + y − 2 = 0.
y = −3x + 2
The slope of the line is the coefficient of the x, so it is equal to -3.
Two parallel lines have the same slope.
f'(a) = 2a − 5
2a − 5 = −3a
a = 1
We find the second coordinate in the function:
y = 1² – 5 · 1 + 6 = 2
So, P=(1, 2)
And then we solve:
y − 2 = −3 (x −1)
y = −3x + 5
We observe that the line is parallel to the one that was given, they have the same slope.
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Taking into account the parabola with vertex V(0, 0), and focus F(0, 2), find the equation of the tangent and of the normal in the point P(4, 2) .
First, we write the equation of the parabola, given that it’s parameter is: p =4.
F(0, 4/2) → F(0,2) → x2 = 2.4y → x2 = 8y
For x = 4 → y = 2 so, P(4,2).
Tangent formula : xox=p(y+yo)
Slope Of Tangent m = 1
Slope of Normal m’ = -1/m
Normal formula: y – yo = m’(x – xo)
For More questions on the tangent and the Normal of the Parabola Read Co- Ordinate Geometry by Arihant Publication.
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