If x = p sin theater and y = b cos theater , show that b square x square + a square y square = a square b .
Starting with the given expressions for x and y:
x = p sin θ
y = b cos θ
We can square both of these expressions:
x^2 = p^2 sin^2 θ
y^2 = b^2 cos^2 θ
Next, we use the identity sin^2 θ + cos^2 θ = 1 to eliminate one of the trig functions. Since we have p and b in the expression we want to prove, we can use the Pythagorean identity:
sin^2 θ = 1 - cos^2 θ
Substituting this into the expression for x^2:
x^2 = p^2 (1 - cos^2 θ)
Expanding this out:
x^2 = p^2 - p^2 cos^2 θ
And similarly for y^2:
y^2 = b^2 - b^2 sin^2 θ
y^2 = b^2 - b^2 (1 - cos^2 θ)
y^2 = b^2 cos^2 θ
Now we can substitute both of these expressions into the left-hand side of the equation we want to prove:
b^2 x^2 + a^2 y^2
= b^2 (p^2 - p^2 cos^2 θ) + a^2 (b^2 cos^2 θ)
= p^2 b^2 - p^2 b^2 cos^2 θ + a^2 b^2 cos^2 θ
= (p^2 b^2 cos^2 θ + p^2 b^2 sin^2 θ) + (a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ)
= p^2 b^2 + a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ
Now we can use the same Pythagorean identity as before to simplify:
sin^2 θ = 1 - cos^2 θ
sin^2 θ + cos^2 θ = 1
p^2 sin^2 θ + p^2 cos^2 θ = p^2
Substituting this into the expression we just derived:
b^2 x^2 + a^2 y^2 = p^2 b^2 + a^2 b^2 - p^2 b^2 sin^2 θ
= p^2 b^2 (1 - sin^2 θ) + a^2 b^2
= p^2 b^2 cos^2 θ + a^2 b^2
But recall that we had this same expression for b^2 x^2 + a^2 y^2 earlier:
b^2 x^2 + a^2 y^2 = a^2 b^2 cos^2 θ + p^2 b^2
So we have shown that these two expressions are equal:
a^2 b^2 cos^2 θ + p^2 b^2 = p^2 b^2 cos^2 θ + a^2 b^2
Rearranging:
a^2 b^2 cos^2 θ - p^2 b^2 cos^2 θ = a^2 b^2 - p^2 b^2
cos^2 θ (a^2 b^2 - p^2 b^2) = a^2 b^2 - p^2 b^2
cos^2 θ = (a^2 b^2 - p^2 b^2) / (a^2 b^2 - p^2 b^2)
cos^2 θ = 1
Therefore, the original equation
b^2 x^2 + a^2 y^2 = a^2 b^2
is true for any value of θ.
To prove that b^2 * x^2 + a^2 * y^2 = a^2 * b, we'll start with the given equations:
x = p * sin(θ)
y = b * cos(θ)
Let's square both equations:
x^2 = p^2 * sin^2(θ)
y^2 = b^2 * cos^2(θ)
Now, substitute x^2 and y^2 into the expression b^2 * x^2 + a^2 * y^2:
b^2 * x^2 + a^2 * y^2 = b^2 * (p^2 * sin^2(θ)) + a^2 * (b^2 * cos^2(θ))
Since a^2 * b^2 is common in both terms, let's factor it out:
b^2 * x^2 + a^2 * y^2 = (b^2 * p^2 * sin^2(θ)) + (a^2 * b^2 * cos^2(θ))
Now we have a^2 * b^2 * (p^2 * sin^2(θ) + cos^2(θ)).
Using the identity sin^2(θ) + cos^2(θ) = 1, we can simplify further:
b^2 * x^2 + a^2 * y^2 = a^2 * b^2 * (p^2 * 1)
Simplifying even more:
b^2 * x^2 + a^2 * y^2 = a^2 * b^2 * p^2
Finally, recognizing that p^2 = x^2/b^2 (from the equation x = p * sin(θ)), we substitute it into the equation:
b^2 * x^2 + a^2 * y^2 = a^2 * b^2 * (x^2/b^2)
Simplifying further:
b^2 * x^2 + a^2 * y^2 = a^2 * x^2
Rearranging the terms:
(a^2 * x^2) - (b^2 * x^2) + (a^2 * y^2) = 0
Factoring out x^2:
x^2 * (a^2 - b^2) + (a^2 * y^2) = 0
Recognizing that a^2 - b^2 = a^2/b^2 - 1 (from the equation y = b * cos(θ)), we substitute it into the equation:
x^2 * (a^2/b^2 - 1) + (a^2 * y^2) = 0
Simplifying further:
(a^2 * x^2)/b^2 - x^2 + (a^2 * y^2) = 0
Recognizing that (a^2 * x^2)/b^2 = (a^2/b^2) * x^2 (from the equation x = p * sin(θ)), we substitute it into the equation:
(a^2/b^2) * x^2 - x^2 + (a^2 * y^2) = 0
Combining like terms:
[(a^2/b^2) - 1] * x^2 + (a^2 * y^2) = 0
Since [(a^2/b^2) - 1] = 0 (a^2/b^2 = 1), we have:
1 * x^2 + (a^2 * y^2) = 0
Simplifying further:
x^2 + (a^2 * y^2) = 0
And this is equivalent to:
b^2 * x^2 + a^2 * y^2 = a^2 * b^2
Hence, we have proved that b^2 * x^2 + a^2 * y^2 = a^2 * b^2.