Question

$( {y}^{2}{a}^{4}) ^{3\2}$
​

Answer 1

Step-by-step explanation:

Given

(x−3)

2

+(y−4)

2

=

9

y

2

(x−3)

2

+(y

2

+16−8y)=

9

y

2

(x−3)

2

+y

2

−

9

y

2

−8y+16=0

(x−3)

2

+

9

8y

2

−8y+16=0

(x−3)

2

+

9

8

(y

2

−9y)=−16[add (

2

9

)

2

]

(x−3)

2

+

9

8

(y

2

−9y+(

2

9

2

))−

9

8

(

2

9

)

2

=−16

(x−3)

2

+

9

8

(y−

2

9

)

2

=−16−1

4

8×9

(x−3)

2

+

9

8

(y−

2

9

)

2

=2

2

(x−3)

2

+

9×2

8

(y−

2

9

)

2

=1

2

(x−3)

2

+

9/4

(y−

2

9

)

2

=1−−−−(1)

Here e

2

=(1−

b

2

a

2

)

So From (1) a

2

=2, b

2

=9/4

e

2

=(1−

9/4

1

)=1−

9

8

=

9

9−8

=

9

1

e

2

=

9

1

e=

3

1

as we know ellipse ecentricity lie between 0<e<1

Answer 2

Answer:

sorry I didn't know answer of this question

Anjali Hooda here