The equation of the curve that passes through (−1, 1) and the slope of line touching to the curve at any point \( (x,y) \) is \( \displaystyle\frac{3x^2}{y} \), is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y^2=2x^3+3 \)
The slope of the tangent is \( \frac{dy}{dx} \), so the curve satisfies a variables separable equation:
\( \displaystyle \frac{dy}{dx}=\frac{3x^2}{y} \)
\( \displaystyle y\,dy=3x^2\,dx \)
\( \displaystyle \frac{y^2}{2}=x^3+C \)
At (−1, 1): \( \frac{1}{2}=-1+C \), so \( C=\frac{3}{2} \).
\( \displaystyle \frac{y^2}{2}=x^3+\frac{3}{2} \)
\( \displaystyle y^2=2x^3+3 \)
Why the other options are incorrect
Option (2) \( y^3=2x^2+3 \) — Incorrect. At (−1, 1) it gives 1 = 5, so the curve does not pass through the point.
Option (3) \( y^2=x^3+\displaystyle\frac{3}{2} \) — Incorrect. This forgets to multiply \( x^3 \) by 2 when clearing the fraction.
Option (4) \( y^2=2x^3+6 \) — Incorrect. The constant is 3, not 6; at (−1, 1) it gives 1 = 4.