The curve passing through the point (−1, 1), given that the slope of the tangent to the curve at any point (x, y) is \(\displaystyle\frac{2x}{y^2}\) also passes through the point \(\left(k, -\displaystyle\frac{1}{2}\right)\), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(8k^2 - 5 = 0\)
The slope condition gives a variables separable equation:
\( \displaystyle \frac{dy}{dx} = \frac{2x}{y^2} \)
\(y^2\,dy = 2x\,dx\)
Integrating both sides:
\( \displaystyle \frac{y^3}{3} = x^2 + C \)
Through (−1, 1):
\( \displaystyle \frac{1}{3} = 1 + C,\quad C = -\frac{2}{3} \)
So the curve is
\(y^3 = 3x^2 - 2\)
At \(\left(k, -\frac{1}{2}\right)\):
\( \displaystyle -\frac{1}{8} = 3k^2 - 2 \)
\( \displaystyle 3k^2 = \frac{15}{8} \)
\( \displaystyle k^2 = \frac{5}{8} \)
\(8k^2 - 5 = 0\)