Math, asked by anamika9713, 1 year ago

A common tangent to the conics x2=6y and 2x24y2=9, is

Answers

Answered by princechauhan8p1dnfc
2

Answer:

Step-by-step explanation:

Find all points where the equations given INTERSECT and have the same SLOPE

... which is the same as saying where the equations share the same tangent line.

Note: dy/dx = ý = slope of an equation.

   x² = 6y            2x² – 4y² = 9  ——>>  y = ± √  [(2x² – 9) ⁄ 4  ]

   2x = 6ý           4x – 8yý = 0

    ý = x ⁄ 3            ý = x ⁄ (2y)

Setting: ý = ý :

         x ⁄ 3 = x ⁄ (2y)

It looks like x=0 is one of the solutions, since the two initial equations appear

to have the same slope at: x=0

First I'll check if the two initial equation even intersect at: x=0 :

  Using the first initial equation: x² = 6y , when: x=0, then: y=0

  Using the second equation: 2x² – 4y² = 9 , when: x=0, then: y=imaginary number

      ... therefore the two equations DO NOT INTERSECT at: x=0

Getting back to the derivative equations:

 1st equation derivative = 2nd equation derivative

         x ⁄ 3 = x ⁄ (2y)  ... recall: (x=0) is NOT a solution   ...   divide by "x"

         1 ⁄ 3 = 1 ⁄ (2y)  ...  where "y" refers to the 2nd initial equation (see above)

           y = 3 ⁄ 2  ... now substitute for "y"

   ± √  [(2x² – 9) ⁄ 4  ] = 3 ⁄ 2

   ± √  [(2x² – 9) ⁄ 4  ] = 3 ⁄ 2

      (2x² – 9) ⁄ 4 = 9 ⁄ 4

       2x² – 9 = 9

         2x² = 18

        x = ± 3  ...  locations where the slopes of each

             ...  equation are equivalent.

Now I'll check to see if the two initial equations intersect at:    x = ± 3

... at:  x = 3 , for equation: x² = 6y ——>> y = 3/2

... and at:  x = 3 , for equation: y = ± √  [(2x² – 9) ⁄ 4  ] ——>> y = ± 3/2

... so they intersect at: (x, y) = (3, 3/2)

   ... at:  x = - 3 , for equation: x² = 6y ——>> y = 3/2

   ... and at:  x = - 3 , for equation: y = ± √  [(2x² – 9) ⁄ 4  ] ——>> y = ± 3/2

   ... so they intersect at: (x, y) = (- 3, 3/2)

Recall the two derivative equations are: (from above):

    ý = x ⁄ 3              ý = x ⁄ (2y)

   ý{- 3} = -3 ⁄ 3             ý{- 3} = -3 ⁄ (2 • (3/2))

   ý{- 3} = -1               ý{- 3} = -1

      ý = x ⁄ 3            ý = x ⁄ (2y)

   ý{3} = 3 ⁄ 3            ý{3} = 3 ⁄ (2 • (3/2))

   ý{3} = 1              ý{3} = 1

Therefore the two initial equations share a common tangent line:

at: (- 3, 3/2) , with the same slope = ý{-3} = -1

and (3, 3/2) , with the same slope = ý{3} = 1

Answered by Anonymous
0

Answer:TAKE IT AS MUCH AS U WANT BUT FAST AS U COULD GOOD BYE

Step-by-step explanation:

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