✨ B R A I N L I E S T Show the necessary steps as you solve for the length of the latus rectum of an ellipse with equation :

3X^2 + Y^2 - 6X + 6Y - 13 = 0

1

:x

2

+y

2

−6x−9y+13=0

(x−3)

2

+(y−

2

9

)

2

−9−

4

81

+13=0

(x−3)

2

+(y−

2

9

)

2

=

4

65

Here,

r

1

=

2

65

C

1

=(3,

2

9

)

Equation of another circle-

S

2

:x

2

+y

2

−2x−16y=0

(x−1)

2

+(y−8)

2

−1−64=0

(x−1)

2

+(y−8)

2

=65

Here,

r

2

=

65

C

2

=(1,8)

Distance between the centre of two circles-

C

1

C

2

=

(3−1)

2

+(8−

2

9

)

2

C

1

C

2

=

4+

4

49

=

2

65

∣r

2

−r

1

∣=

∣

∣

∣

∣

∣

∣

65

−

2

65

∣

∣

∣

∣

∣

∣

=

2

65

∵C

1

C

2

=∣r

1

−r

2

∣

Thus the two circles touches each other internally.

Since the circle touches each other internally. The point of contact P divides C

1

C

2

externally in the ratio r

1

:r

2

, i.e.,

2

65

:

65

=1:2

Therefore, coordinates of P are-

⎝

⎜

⎜

⎜

⎜

⎜

⎛

1−2

1(1)−2(3)

,

1−2

1(8)−2(

2

9

)

⎠

⎟

⎟

⎟

⎟

⎟

⎞

=(5,1)

Therefore,

Equation of common tangent is-

S

1

−S

2

=0

(5x+y−6(

2

x+5

)−9(

2

y+1

)+13)−(5x+y−2(

2

x+5

)−16(

2

y+1

))=0

2

−6x−9y−13

+x+8y+13=0

4x−7y−13=0

Hence the point of contact is (5,1) and the equation of common tangent is 4x−7y−13=0.

Step-by-step explanation:

3x^2+y^2-6x+6y-13=0

= 3x^2+y^2-6x+6y=13p

step-by-step explanation:

the answer is 128