Question

identify the type of conic and find centre ,foci ,vertices and directrix of x2-4y2+6x+16y-11=0 ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The given equation represents a hyperbola.

Center of the hyperbola is (-3, 2)

Focii of the hyperbola are (-3 + √5, 2) and (-3 - √5, 2)

Vertices of the hyperbola are (-1, 2) and (-5, 2)

Directrix are: y =  2x and y = -2x

Step-by-step explanation:

Given equation : x² - 4y² + 6x + 16y - 11 = 0

=> x² - 4y² + 6x + 16y  = 11

=> x² + 6x - [4y² - 16y]  = 11

=>  x² + 2*3*x + 9 - 9 - [(2y)² - 2*2y*4 + 16 - 16]  = 11

=>  (x+3)²  - 9 - [(2y-4)² - 16]  = 11

=> (x+3)²  - 9 -(2y-4)² + 16   = 11

=> (x+3)²  - (2y-4)²    = 11 - 16 + 9

=>  (x+3)² - (2y-4)²  = 4

=> $\frac{(x+3)^{2}}{4}$ - $\frac{(2y-4)^{2}}{4}$ = 1

=>  $\frac{(x+3)^{2}}{2^{2}}$ - $\frac{[2(y-2)]^{2}}{2^{2}}$ = 1

=> $\frac{(x+3)^{2}}{2^{2}}$ - $\frac{2^{2}(y-2)^{2}}{2^{2}}$ = 1

=> $\frac{(x+3)^{2}}{2^{2}}$ - (y-2)²   = 1

We know that the equation of a hyperbola is  $\frac{(x-h)^{2}}{a^{2}}$ - $\frac{(y-k)^{2}}{b^{2}}$ = 1

with center = (h, k)

           focii = (h ± c, k)           where c² = a² + b²

     vertices = (h ± a, k)

 directrix = y = ± ax/b

Therefore, here,

The equation represents a hyperbola

Center of the hyperbola = (-3, 2)

c² = a² + b²

=> c² = 2² + 1² = 4 + 1 = 5

=> c = √5

Focii of the hyperbola = (-3 ± √5, 2)

                                     = (-3 + √5, 2) and (-3 - √5, 2)

Vertices of the hyperbola = (-3 ± 2, 2)

                                     = (-3 + 2, 2) and (-3 - 2, 2)

                                     = (-1, 2) and (-5, 2)

Directrix => y = ± 2x/1

                   y = ± 2x

                   y =  2x and y = -2x