AP EAMCET · Maths · Application of Derivatives
If the tangent drawn to the curve \(\left(x^2+1\right)(y-3)=x\) at a point \(\mathrm{P}\), lying in the first quadrant, is a horizontal line, then the equation of the normal at the point \(\mathrm{P}\) is
- A \(x=\frac{7}{2}\)
- B \(x=1\)
- C \(y=\frac{7}{2}\)
- D \(y=1\)
Answer & Solution
Correct Answer
(B) \(x=1\)
Step-by-step Solution
Detailed explanation
\(\because\left(x^2+1\right)(y-3)=x\)... (i) \(\begin{aligned} & \text { Then, }\left(x^2+1\right) \frac{d y}{d x}+(2 x+0)(y-3)=1 \\ & \Rightarrow \frac{d y}{d x}=\frac{6 x-2 x y+1}{x^2+1}\end{aligned}\) \(\because\) Tangent is a horizontal line.…
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