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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Conic Sections in Trigonometry

Conic Sections in Trigonometry

In this article we are going to learn about conic sections i.e. circles, hyperbola, parabola, ellipse etc. All these are formed by the intersection of plane and cone.

Table of Content
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Conic sections or conics are used to analyze circles, ellipses, hyperbolas, and parabolas since they may be obtained by intersecting a plane with a right circular cone. These curves have a wide range of uses, and we’ve utilized them in a variety of studies. In terms of the following, all conics can be written: Ax2 + Bxy + Cy2 + Dx + Ey + F. Now, let us understand each concept of conic section in- depth.

Sections of a cone 

If we take an intersection of a plane with a cone then the interaction is known as the conic section. We obtained different types of conic sections depending on the position of the intersecting plane with the cone and the angle made by it with a vertical axis of the cone.

The conics in which planes intersect the vertex of the cone then this is known as degenerate conic else non-degenerate conic.

  1. When β= 90°The section is a circle.

  2. When α<β<90°, the section is an ellipse 

  3. When β=α; the section is a parabola 

  4. When 0 ≤β < α; the section is hyperbola

  5. When α<β≤90°, then the section is a point 

  6. When β = α, the plane contains a generator of the cone and the section is a straight line 

Circles

Circles are defined as all locations in a plain that are equally spaced from a fixed point. The radius of the circle is the distance between the fixed point and the radius of the circle is the distance between the fixed point and the fixed point.

Consider a circle of radius r. Let the center of the circle be C(h,k) and P(x,y) be some point on the circumference. We know that the distance from center to point P is known as radius of a circle so it is equals to r, by the distance formula 

√(x-h)2+(y-k)2 = r

(x-h)2+(y-k)2= r2

This is known as the equation of a circle with center (h,K) and radius r.

Ellipse 

It is the set of all points in a plane, the sum of whose distance from two fixed points in the plane is constant.

  1. Center of ellipse – The midpoint joining the foci 

  2. Major axis – the line segment through the foci of the ellipse. It is denoted by 2a 

  3. Minor axis – line segment through the center and perpendicular to the major axis. It is denoted by 2b.

  4. Vertices – the end points of the major axis.

Relationship between semi-major axis, semi-minor axis and the distance of the focus from the center of the ellipse.

Take a point p on the one end of major axis

Sum of distances of the point P to the foci is F1P + F2P = F1O +OP + F2P

Since, F1P = F1O + OP

c+a+a-c=2a

take another point Q on the one end of the minor axis 

sum of distance of the point Q 

F1Q + F2Q = √b2+c2 + √b2+c2 = 2 √b2+c2

Since both P and Q lies on the ellipse 

2√b2+c2 = 2a;    a=√b2+c2

a2=b2+c2;      c= √a2-b2

standard equation of ellipse : 

x2/a2+y2/b2=1

Parabola 

A parabola is a set of all points that are equidistant from a fixed line called directrix and fixed point called focus.

A line through focus and perpendicular to the directrix is known as the axis of parabola.

The straight line formed if the fixed point lies on the fixed line, which is equidistant from the fixed point and perpendicular to the fixed line then it is known as the degenerate case of parabola.

To derive the equation of parabola 

Let us take F as focus 

l- directrix

FM- perpendicular to the directrix and bisects at point O.

Now produce MOto X 

The mid-point O is called the vertex of the parabola. Let OX- x-axis and OY – y-axis.

Distance from directrix from focus is 2a and the coordinates of the focus are (a,0) then, equation of directrix= x+a = 0 

Let P(x,y) be any point on a parabola such that 

PF=PB

Where PB is perpendicular to l. 

The coordinates of B(-a,y) . by the distance formula 

PF = √(x-a)2+y2

PB = √(x+a)2

Since, PF=PB

√(x-a)2+y =√(x+a)2

(x-a)2+y2 =x+a2

x2-2ax+ a2 + y2 =x2+2ax+a2

y2 = 4ax (a>0) 

Hence, the equation of parabola is

y2 = 4ax

Latus Rectum 

The line perpendicular to the axis of the parabola is known as the latus rectum of a parabola. 

To find the length of latus rectum of the parabola
y2 = 4ax

By definition, AF=AC

But AC=FM=2a 

AF=2a 

As we know that parabola is symmetric to x-axis AF=FB and so AB= length of the latus rectum = 4a

Hyperbola 

A hyperbola is a set of all points in a plane, the difference of whose distances from two fixed points in the plane is constant.

The standard equation of hyperbola is:

  1. The equation of hyperbola with origin (0,0) and transverse axis along x-axis is x2/a2-y2/b2=1 

Conclusion

A circle is the set of all points in a plane that are equidistant from a plane. The equation of a circle from center (h,k) and the radius r is (x-h)2+(y-k)2= r2 . The latus rectum of a parabola is a line segment perpendicular to the axis of the parabola. The length of latus rectum of parabola is y2 = 4ax is 4a.

Standard equation for non-degenerate conic section: Circle – x2+y2=r2,

Ellipse – x2/a2+y2/b2=1

Parabola- y2 -4ax=0,

Hyperbola- x2/a2-y2/b2=1 

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Find the center and the radius of the circle x2+y2+8x+10y-8=0

Ans : x2...Read full

Find the equation of the parabola with focus (2,0) and directrix x=-2.

Ans : Since the focus lies on the x-axis the x- axis itself is the axis of the parabola. Since the directrix is (2,0...Read full

Ans :

x2+8x+(y2+10y)=8

x2 +2.4.x+y2+2.5.y=8

[x2 +2.4.x+ 42 -42] +y2+2.5.y+ 52-52=8

(x2+8x+16)+y2+10y+25=8+16+25

(x+4)2+(y+5)2=49

{x-(-4)}2 + {y-(-5)}2 = 72 ……1

Comparing equation 1 with (x-h)2 +(y-k)2 = r2 

h= -4, k = -5, r=7

thus, center of circle = (h,K) 

= (-4,-5) and radius (r) = 7

Ans : Since the focus lies on the x-axis the x- axis itself is the axis of the parabola. Since the directrix is (2,0) the parabola is to be y2 =4ax with a=2. Hence required equation is 

y2 = 4(2)x = 8x.

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