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A conic is an degenarate curves coming out as a result of intersection of plane with one and two Nappies. For the plane perpendicular to the axis of the cone a circle is generated. On a plane a parabola or ellipse is created. On double planes of a cone a hyperbola is created.

The curve of intersection of a plane with a cone depends on the inclination of the axis of the cone to the cutting plane.

If the cutting plane is parallel to one and only one generator then the conic is a parabola. If the cutting plane is parallel to no generator, then the conic is an ellipse.

When the cutting plane is parallel to two generators, then it intersects both nappes of the cone and the conic is called a

In geometry, conic section is the locus of a point which moves so that its distance from a fixed point bears a constant ratio to its distance from a fixed line.

Consider the following figure.

In the figure,$\frac{SP}{PM}$ = Constant=e $\Rightarrow$ SP = e PM

The fixed point(S, S’) is called the focus of the conic and this fixed line is called the directrix of the conic. Also this constant ratio is called the eccentricity of the conic and is denoted by e.

If e < 1, the conic is called Ellipse.

If e > 1, the conic is called Hyperbola.

If e = 0, the conic is called circle.

If e = ∞ , the conic is called pair of straight lines.

∆ = abc + 2fgh – af

ax

In this equation (1) represents the Degenerate conic whose nature is given in the following table.

Condition |
Nature of Conic |

∆ = 0 and ab – h^{2 }= 0 |
A pair of straight lines or empty set. |

∆ = 0 and ab – h^{2} ≠ 0 |
A pair of intersecting straight lines. |

∆ = 0 and ab – h^{2} < 0 |
Real or imaginary pair of straight lines |

∆ = 0 and ab – h^{2} > 0 |
Point |

In this case equation (1) represents the Non-degenerate conic whose nature is given in the following table:

Condition |
Nature of Conic |

∆ ≠ 0, h=0, a=b | A circle |

∆ ≠ 0, ab – h^{2} = 0 |
A parabola |

∆ ≠ 0, ab – h^{2} > 0 |
An Ellipse or empty set |

∆ ≠ 0, ab – h^{2} < 0 |
A hyperbola |

∆ ≠ 0, ab – h^{2} < 0 and a+b=0 |
A rectangular hyperbola |

∆ = abc + 2fgh – af

The above equation will become,

ax

Radius r = $\sqrt{(\frac{-g}{a})^{2}+(\frac{-f}{a})^{2}-C}$

Comparing this with the second degree equation, ax

we observe that a = b = 1, and h = 0.

From the table it is clear that this represents the equation of circle.

Let us evaluate, ∆ = abc + 2fgh – af

= -8 + 0 -4 - 9 – 0

= -21 ≠ 0.

Hence the above system of equation represent the equation of a circle

whose centre is ( -g, -f) = (3, -2) and radius = $\sqrt{9+4-(-8)}$

=

∆ = abc + 2fgh – af

comparing this with the second degree equation, ax

we get, a= 13, b= 37,c=-2

2h = -18 => h = -9

2g = 2 => g = 1

2f = 14 => f= 7

Therefore, ∆ = abc + 2fgh – af

= -962 -126 -637 -37 + 162

= - 1600 ≠ 0

and also ab – h

Hence the conditions are satisfied as per the above table for an ellipse.

Therefore the given equation represents equation of an ellipse.

The general, the equation of an ellipse is, $\frac{(x-h)^{2}}{a^{2}}$ + $\frac{(y-k)^{2}}{b^{2}}$ = 1, where a > b or a < b.

If a= b, the above equation becomes the equation of a circle.

Let us consider the equation, $\frac{(x-h)^{2}}{a^{2}}$ + $\frac{(y-k)^{2}}{b^{2}}$ = 1,where a>b.

The co-ordinates of its center is ( h, k)

The Major axis is x-axis and Minor axis is y-axis.

Length of the major axis is 2a and length of the minor axis is 2b.

The co-ordinates of the vertices are ( ± a, 0 ), ( 0, ± b)

The co-ordinates of foci are ( (ae+h, k) and (-ae+h, k)

The equation of directrices are x = $\frac{a}{e}$ + h and x = - $\frac{a}{e}$ + h

The eccentricity is calculated using the relation b

Length of the latus rectum is $\frac{2b^{2}}{a}$

$\Rightarrow$ $\frac{16x^{2}+25y^{2}}{1600}$ = $\frac{160}{1600}$ =1

$\Rightarrow$ $\frac{x^{2}}{100}$ + $\frac{y^{2}}{64}$ = 1

$\Rightarrow$ a

$\Rightarrow$ a = ± 10 , and b = ± 8, since a > b, the equation represents the ellipse whose major axis is horizontal.

Since h = k = 0, The center of the ellipse is (0, 0 )

We have, b

$\Rightarrow$ 64 = 100( 1 – e

$\Rightarrow$ 64 = 100 – 100 e

$\Rightarrow$ 100 e

$\Rightarrow$ $e^{2}$ =$\frac{36}{100}$

$\Rightarrow$ $e$ = $\frac{6}{10}$ = $\frac{3}{5}$ = 0.6< 1

$\Rightarrow$ e = 0.6 or $\frac{3}{5}$ The co-ordinates of foci will be, ( ± ae , 0)

$\Rightarrow$ (± (10)$\frac{3}{5}$,0)

$\Rightarrow$ (± 6 , 0)

The equation of directrices are x = ± $\frac{a}{e}$ = ± $\frac{10}{0.6}$ = ± $\frac{50}{3}$

The ellipse can be drawn by marking, the vertices, center, focal points. The directrices are drawn parallel to the x-axis at a distance of $\frac{50}{-3}$ units [ which is 16.67 units ]

The graph of the Ellipse is shown below.

Question 2:Find the parametric equation of conic section, the equation is x

Question 3:What does the conic x

Question 4:What is the directrix for the parabola (y+1)