The tangent to the curve y2 + 2x − 5 = 0 at the point (h, k) is paral…

The correct answer is Option (3) → 2

Differentiating \(y^2 + 2x - 5 = 0\) with respect to x:

    \( \displaystyle 2y\frac{dy}{dx} + 2 = 0 \)

    \( \displaystyle \frac{dy}{dx} = -\frac{1}{y} \)

So the slope of the tangent at (h, k) is \(-\frac{1}{k}\).

The line x + 2y = 4 can be written \(y = -\frac{1}{2}x + 2\), so its slope is \(-\frac{1}{2}\).

Parallel lines have equal slopes:

    \( \displaystyle -\frac{1}{k} = -\frac{1}{2} \)

    \(k = 2\)

The point is on the curve: \(4 + 2h - 5 = 0\) gives \(h = \frac{1}{2}\), so (h, k) = \(\left(\frac{1}{2}, 2\right)\).