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: Understanding the Concept
If a line touches a curve at a point, two things hold: the point lies on the curve, and the slope of the curve at that point equals the slope of the line.

Step 2: Use the point
\((1,-1)\) is on \(y^2=ax^3+b\):
\[ 1=a+b \]

Step 3: Use the slope
The line \(x+y=0\) has slope \(-1\). Differentiate the curve:
\[ 2y\frac{dy}{dx}=3ax^2\Rightarrow\frac{dy}{dx}=\frac{3ax^2}{2y} \]
At \((1,-1)\):
\[ \frac{3a}{-2}=-1\Rightarrow a=\frac23 \]

Step 4: Find b
\[ b=1-a=1-\frac23=\frac13 \]
So \(a=\dfrac23\) and \(b=\dfrac13\), option (C). Option (B) has the values swapped.

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