Find the value of k for which the point (k-1, k) lies inside the parabola \(y^2\) = 4x.

Solution :

\(\because\) Point (k-1, k) lies inside the parabola \(y^2\) = 4x.

\(\therefore\)  \({y_1}^2 – 4ax_1\) < 0

\(\implies\)  \(k^2\) – 4(k-1) < 0

\(\implies\)  \(k^2\) – 4k + 4 < 0

\((k-2)^2\) < 0 \(\implies\) k \(\in\) \(\phi\)

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