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3 years ago

Foci of ellipse of 3x2 + y2 =9

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

8 0

Answer:

x=√-1/3 y^2 +3 or x=-√-1/3 y^2 +3

Step-by-step explanation:

Let's solve for x.

3x2+y2=9

Step 1: Add -y^2 to both sides.

3x2+y2+−y2=9+−y2

3x2=−y2+9

Step 2: Divide both sides by 3.

3x^2/3= -y^2+9/3

x^2=-1/3 y^2 +3

x=√-1/3 y^2 +3 or x=-√-1/3 y^2 +3

Hope it helped!

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BC = 36; D = 15. What is the value of BD - AC.<br> B<br> A<br> BD - AC =

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Answer:

BD - AC = 0

Step-by-step explanation:

Given that:

BC = 36 , CD = 15 and BD = x

As these are forming right angled triangle,

Using Pythagorean theorem,

$(BD)^2+(CD)^2=x^2\\(36)^2+(15)^2 = x^2\\1296 + 225 = x^2\\1521 = x^2\\$

Taking square root on both sides

$\sqrt{x^2}=\sqrt{1521}\\x=39$

As the given shape is rectangular, the sides parallel to each other will have same values.

AB = 15 , AD = 36 and AC = x

$(15)^2+(36)^2 = x^2 \\225 + 1296 = x^2 \\1521 = x^2 \\x = 39$

Now,

BD - AC = 39 - 39 = 0

Hence,

BD - AC = 0

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