FINDING CENTER FOCI VERTICES AND DIRECTRIX OF ELLIPSE AND HYPERBOLA

About "Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola"

Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola :

Here we are going to see some example problems to understand the concept of finding vertex, focus length of latus rectum and equation of directrix of ellipse and hyperbola.

Details of ellipse whose center is (0, 0)

Symmetric about x- axis

Center :

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- axis

Center :

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)

Note :

e = √[1 + (b²/a²)]

b²= a²(e² - 1)

Details of ellipse whose center is (0, 0)

Symmetric about x- axis

Center :

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 = ± (a/e)

Symmetric about y- axis

Center :

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 = ± (a/e)

Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola - Practice questions

Question 1 :

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

(i) (x²/25) + (y²/9) = 1

Solution :

The given conic represents the " Ellipse "

The given ellipse is symmetric about x - axis.

a² = 25 and b² = 9

Center :

C (0, 0)

Foci :

F₁ (ae, 0) F₂ (-ae, 0)

e = √1- (b²/a²)

e = √1- (9/25)

e = √(16/25)

e = 4/5

ae = 5(4/5) = 4

F₁ (4, 0) F₂ (-4, 0)

Vertices :

A (a, 0) A' (-a, 0)

A (5, 0) A' (-5, 0)

Equation of directrices :

x = ± a/e

x = ± 5/(4/5)

x = ± 25/4

Let us look into the next example problem on "Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola".

(ii) (x²/3) + (y²/10) = 1

Solution :

The given conic represents the " Ellipse "

The given ellipse is symmetric about y - axis.

a² = 10 and b² = 3

Center :

C (0, 0)

Foci :

F₁ (ae, 0) F₂ (-ae, 0)

c² = a² - b²

c² = 10 - 3

c = √7

e = c/a ==> √(7/10)

ae = √10√(7/10) = √7

F₁ (0, √7) F₂ (0, -√7)

Vertices :

A (a, 0) A' (-a, 0)

A (0, √10) A' (0, -√10)

Equation of directrices :

y = ± a/e

y = ± √10/√(7/10)

y = ± 10/√7

Let us look into the next example problem on "Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola".

(iii) (x²/25) - (y²/144) = 1

Solution :

The given conic represents the " Hyperbola "

The given ellipse is symmetric about x - axis.

a² = 25 and b² = 144

Center :

C (0, 0)

Foci :

F₁ (ae, 0) F₂ (-ae, 0)

c² = a² + b²

c² = 25 + 144

c² = 169

c = 13

e = c/a = 13/5

ae = 5(13/5) = 13

F₁ (13, 0) F₂ (-13, 0)

Vertices :

A (a, 0) A' (-a, 0)

A (5, 0) A' (-5, 0)

Equation of directrices :

x = ± a/e

x = ± 5/(13/5)

x = ± 25/13

(iv) (y²/16) - (x²/9) = 1

Solution :

The given conic represents the " Hyperbola "

The given ellipse is symmetric about y - axis.

a² = 16 and b² = 9

Center :

C (0, 0)

Foci :

F₁ (ae, 0) F₂ (-ae, 0)

c² = a² + b²

c² = 16 + 9

c² = 25

c = 5

e = c/a = 5/4

ae = 4(5/4) = 5

F₁ (5, 0) F₂ (-5, 0)

Vertices :

A (a, 0) A' (-a, 0)

A (4, 0) A' (-4, 0)

Equation of directrices :

x = ± a/e

x = ± 4/(5/4)

x = ± 16/5

After having gone through the stuff given above, we hope that the students would have understood, "Finding Center Foci Vertices and Directrix of Ellipse and Hyperbola".

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