# IDENTIFYING THE TYPE OF CONIC AND FIND FOCI VERTICES AND DIRECTRICES

## About "Identifying the Type of Conic and Find Foci Vertices and Directrices"

Identifying the Type of Conic and Find Foci Vertices and Directrices :

Here we are going to see some example problems to identifying the type of conic and find foci vertices and directrices.

## Details of ellipse whose center is (0, 0)

 Symmetric about x- axisCenter : C (0, 0)Foci : F1 (ae, 0) F2 (-ae, 0)Vertices on major axis  :A (a, 0) A' (-a, 0)Vertices on minor axis  : B (0, b) B' (0, -b)Equation of directrices :x  =  ± (a/e) Symmetric about y- axisCenter :C (0, 0)Foci :F1 (0, ae) F2 (0, -ae)Vertices on major axis  :A (0, a) A' (0, -a)Vertices on minor axis  :B (b, 0) B' (-b, 0)Equation of directrices :y  =  ± (a/e)

## Details of ellipse whose center is (0, 0)

 Symmetric about x- axisCenter : C (h, k)Foci :F1 (h − c, k ) F2 (h + c, k )Vertices on major axis  :A (h -a, k) A' (h + a, k)Equation of directrices :x  =  h ± (a/e) Symmetric about y- axisCenter :C (h, k)Foci :F1 (h, k - c) F2 (h, k + c)Vertices on major axis  :A (h, k - a ) A' (h, k + a)Equation of directrices :y  = k ± (a/e)

Note :

e  =  √[1 + (b2/a2)]

b=  a2(e2 - 1)

## Identifying the Type of Conic and Find Foci Vertices and Directrices - Practice questions

Question 1 :

Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

(i)  [(x - 3)2/225] + [(y - 4)2/289] = 1

Solution :

The given conic represents the " Ellipse "

The given ellipse is symmetric about y - axis.

a2  =  289 and b =  225

a  =  17 and b  =  15

c2  =  a2 - b2

c2  =  289 - 225

c2  =  64

c  =  8

e = c/a  =  8/17

Center :

C (h, k)  ==>  (3, 4)

Foci :

F1 (h, k - c ) F2 (h, k + c )

F1 (3, 4 - 8 ) F2 (3, 8 + 4 )

F1 (3, -4 ) F2 (3, 12 )

Vertices :

A (h, k - a ) A' (h, k + a)

A (3, 4 - 17 ) A' (3, 4 + 17)

A (3, -13 ) A' (3, 21)

Equation of directrices :

y  =  k ± (a/e)

y  =  4 ± (17/(8/17))

y  =  4 ± (289/8)

 y  =  4 + (289/8)   y  =  (32+289)/8  y   =  321/8 y  =  4 - (289/8)   y  =  (32 - 289)/8  y  =  - 257/8

(ii)  [(x + 1)2/100] + [(y - 2)2/64] = 1

Solution :

The given conic represents the " Ellipse "

The given ellipse is symmetric about x - axis.

a2  =  100 and b =  64

a  =  10 and b  =  8

c2  =  a2 - b2

c2  =  100 - 64

c2  =  36

c  =  6

e  =  c/a  =  6/10  =  3/5

Center :

C (h, k)  ==>  (-1, 2)

Foci :

F1 (h - c, k) F2 (h + c, k)

F1 (-1-6, 2) F2 (-1 + 6, 2)

F1 (-7, 2) F2 (5, 2)

Vertices :

A (h - a, k) A' (h + a, k)

A (-1 - 10, 2) A' (-1 + 10, 2)

A (-11, 2) A' (9, 2)

Equation of directrices :

x  =  h ± (a/e)

x  =  -1 ± (10/(3/5))

x  =  -1 ± (50/3)

 x  =  -1 + (50/3)  x  =  (-3+50)/3  x   =  47/3 x  =  -1 - (50/3)  x  =  (-3-50)/3  x   =  -53/3  After having gone through the stuff given above, we hope that the students would have understood, "Identifying the Type of Conic and Find Foci Vertices and Directrices".

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