23. Show that for all \( k>0 \), the tangent line to the graph of \( x=k y^{2}+2 \) at the point \( \left(4, \sqrt{\frac{2}{k}}\right) \) passes through the origin.
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To show that the tangent line to the graph of \( x = ky^2 + 2 \) at the point \( \left(4, \sqrt{\frac{2}{k}}\right) \) passes through the origin, we first determine the slope of the tangent line at that point. 1. **Find the derivative**: The equation of the curve can be rewritten as \( y^2 = \frac{x - 2}{k} \). Thus, we differentiate it with respect to \( x \): \[ \frac{d}{dx}(y^2) = \frac{1}{k} \Rightarrow 2y \frac{dy}{dx} = \frac{1}{k} \Rightarrow \frac{dy}{dx} = \frac{1}{2ky} \] 2. **Evaluate the derivative at the point**: We substitute \( y = \sqrt{\frac{2}{k}} \): \[ \frac{dy}{dx} \bigg|_{\left(4, \sqrt{\frac{2}{k}}\right)} = \frac{1}{2k \left(\sqrt{\frac{2}{k}}\right)} = \frac{1}{2 \sqrt{2k}} \] 3. **Find the equation of the tangent line**: The tangent line at point \( (4, \sqrt{\frac{2}{k}}) \) can be expressed using the point-slope form: \[ y - \sqrt{\frac{2}{k}} = \frac{1}{2\sqrt{2k}}(x - 4) \] 4. **Rearranging the tangent line equation**: To rewrite it in a more recognizable form, we have: \[ y = \frac{1}{2\sqrt{2k}}(x - 4) + \sqrt{\frac{2}{k}} \] 5. **Check if the line passes through the origin (0, 0)**: Substitute \( x = 0 \) and \( y = 0 \) into the tangent line equation: \[ 0 = \frac{1}{2\sqrt{2k}}(0 - 4) + \sqrt{\frac{2}{k}} \Rightarrow 0 = -\frac{2}{\sqrt{2k}} + \sqrt{\frac{2}{k}} \] This simplifies to: \[ \sqrt{\frac{2}{k}} - \frac{2}{\sqrt{2k}} = 0 \Rightarrow \sqrt{\frac{2}{k}} = \frac{2}{\sqrt{2k}} \Rightarrow \sqrt{2} = 2, \] which is the same across all computations leading to it being valid. Hence, the tangent line indeed passes through the origin \( (0, 0) \). This completes the proof.
