MHT CET · Maths · Application of Derivatives
\(\mathrm{A}(1,-3), \mathrm{B}(4,3)\) are two points on the curve \(y=x-\frac{4}{x}\). The points on the curve, the tangents at which are parallel to the chord \(\mathrm{AB}\), are
- A \((1,2),(-1,-2)\)
- B \((2,0),(-2,0)\)
- C \((0,2),(1,-2)\)
- D \((3,2),(-3,1)\)
Answer & Solution
Correct Answer
(B) \((2,0),(-2,0)\)
Step-by-step Solution
Detailed explanation
Slope of tangent \(=\) slope of \(\mathrm{AB}=\frac{3+3}{4-1}=\frac{6}{3}=2\)
\(y=x-\frac{4}{x}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{4}{x^2}\)
\(\begin{aligned} & \Rightarrow 2=1+\frac{4}{x^2} \\ & \Rightarrow x^2=4 \\ & \Rightarrow x= \pm 2\end{aligned}\)
When \(x=2, y=2-\frac{4}{2}=0\)
When \(x=-2, y=-2+\frac{4}{2}=0\)
\(\therefore \quad\) The required points are \((2,0)\) and \((-2,0)\).
\(y=x-\frac{4}{x}\)
\(\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\frac{4}{x^2}\)
\(\begin{aligned} & \Rightarrow 2=1+\frac{4}{x^2} \\ & \Rightarrow x^2=4 \\ & \Rightarrow x= \pm 2\end{aligned}\)
When \(x=2, y=2-\frac{4}{2}=0\)
When \(x=-2, y=-2+\frac{4}{2}=0\)
\(\therefore \quad\) The required points are \((2,0)\) and \((-2,0)\).
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