AP EAMCET · Maths · Application of Derivatives
The curve \(3 y^2=2 a x^2+6 b\) passes through the point \(P(3,-1)\) and the gradient of the curve at \(P\) is " -1 ", then the values of \(a\) and \(b\) are
- A \(a=\frac{1}{2}, b=-1\)
- B \(a=\frac{-1}{2}, b=1\)
- C \(a=\frac{1}{2}, b=1\)
- D \(a=\frac{-1}{2}, b=-1\)
Answer & Solution
Correct Answer
(A) \(a=\frac{1}{2}, b=-1\)
Step-by-step Solution
Detailed explanation
Equation of given curve is \(3 y^2=2 a x^2+6 b\)...(i) \(\because\) Curve (i) passes through point \(P(3,-1)\), so \(3=18 a+6 b\) \(\Rightarrow \quad 6 a+2 b=1\)...(ii) and the \(\left.\frac{d y}{d x} \right\rvert\,(3,-1)=-1\) (given), so on differentiating the Eq. (i), we get…
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