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Implicit Differentiation

If x,y ∈ R ...

Question

If $x, y \in R$ and $2 x^{2} + 6 x y + 5 y^{2} = 1$ , then

A

| x | \leq \sqrt{5}

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B

| x | \geq \sqrt{5}

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C

y^{2} \leq 2

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D

y^{2} \leq 4

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Solution

The correct options are
A $| x | \leq \sqrt{5}$
C $y^{2} \leq 2$

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Similar questions

x, y \in R

2 x^{2} + 6 x y + 5 y^{2} = 1,

x = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}

y = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}

, find the value of

x^{2} + y^{2} - 6 x y

P = 7 x^{2} + 5 x y - 9 y^{2}, Q = 4 y^{2} - 3 x^{2}

R = 4 x^{2} + x y + 5 y^{2}

P + Q + R = 8 x^{2} + 6 x y

x, y \in R

satisfy the equation

x^{2} + y^{2} - 4 x - 2 y + 5 = 0

then the value of the expression

\frac{[(\sqrt{x} - \sqrt{y})^{2} + 4 \sqrt{x y})]}{(x + \sqrt{x y})}

x, y \in R

, satisfy the equation

x^{2} + y^{2} - 4 x - 2 y + 5 = 0

, then the value of the expression

\frac{(\sqrt{x} - \sqrt{y})^{2} + 4 \sqrt{x y}}{x + \sqrt{x y}}

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