Let $$ x=k y^{2}+2 $$, where $$ k0 $$. 22. Show that for all $$ k0 $$, the point $$ \left(4, \sqrt{\frac{2}{k}}\right) $$ is on the graph of $$ x=k y^{2}+2 $$.

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Let $$ x=k y^{2}+2 $$, where $$ k>0 $$. 22. Show that for all $$ k>0 $$, the point $$ \left(4, \sqrt{\frac{2}{k}}\right) $$ is on the graph of $$ x=k y^{2}+2 $$.

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To show that the point $$ \left(4, \sqrt{\frac{2}{k}}\right) $$ is on the graph of the equation $$ x = k y^2 + 2 $$, we need to verify that when we substitute $$ x = 4 $$ and $$ y = \sqrt{\frac{2}{k}} $$ into the equation, both sides are equal.

1. Start with the equation:
   $$
   x = k y^2 + 2
   $$

2. Substitute $$ x = 4 $$ and $$ y = \sqrt{\frac{2}{k}} $$:
   $$
   4 = k \left(\sqrt{\frac{2}{k}}\right)^2 + 2
   $$

3. Calculate $$ \left(\sqrt{\frac{2}{k}}\right)^2 $$:
   $$
   \left(\sqrt{\frac{2}{k}}\right)^2 = \frac{2}{k}
   $$

4. Substitute this back into the equation:
   $$
   4 = k \cdot \frac{2}{k} + 2
   $$

5. Simplify the right side:
   $$
   4 = 2 + 2
   $$
   $$
   4 = 4
   $$

Since both sides of the equation are equal, we have shown that the point $$ \left(4, \sqrt{\frac{2}{k}}\right) $$ is indeed on the graph of $$ x = k y^2 + 2 $$ for all $$ k > 0 $$.
To show that the point $$ \left(4, \sqrt{\frac{2}{k}}\right) $$ is on the graph of $$ x = k y^2 + 2 $$, substitute $$ x = 4 $$ and $$ y = \sqrt{\frac{2}{k}} $$ into the equation. This results in both sides being equal, confirming the point lies on the graph for all $$ k > 0 $$.

Let \( x=k y^{2}+2 \), where \( k>0 \). 22. Show that for all \( k>0 \), the point \( \left(4, \sqrt{\frac{2}{k}}\right) \) is on the graph of \( x=k y^{2}+2 \).

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