Question

The line \(x+y = 0\) touches the curve \(y^2 = ax^3+b\) at \((1,-1)\) then values of \(a\) and \(b\) respectively are ...........

• A. \(\frac{1}{2},\frac{2}{5}\)
• B. \(\frac{1}{3},\frac{2}{3}\)
• C. \(\frac{2}{3},\frac{1}{3}\)
• D. \(\frac{2}{5},\frac{1}{2}\)

Hint:

The point lies on the curve and the slope of the curve at that point equals the slope of the line.

Answer 1

The Correct Option is C

Step 1: Approach
Treat it as a system of two equations in $a$ and $b$.

Step 2: Equation 1
Passing through $(1,-1)$ gives $a+b=1$.

Step 3: Equation 2
Implicit differentiation gives $y'=\dfrac{3ax^2}{2y}$, which equals $-\dfrac{3a}{2}$ at the point. Setting it equal to the slope of $x+y=0$, which is $-1$, gives $a=\dfrac23$.

Step 4: Solution
Then $b=1-\dfrac23=\dfrac13$. This is option (C), and it satisfies both equations.

Final Answer:
The curve passes through (1, -1) and has slope -1 there, giving a = 2/3 and b = 1/3, option (C). \[ \boxed{a=\frac23,\ b=\frac13} \]