AP EAMCET · Maths · Application of Derivatives
If the tangent drawn at \(A(2,1)\) to the curve \(\mathrm{x}=1+\frac{1}{\mathrm{y}^2}\) meets the curve again at \(\mathrm{B}\), then
- A the tangent drawn at \(\mathrm{B}\) coincides with the tangent drawn at \(\mathrm{A}\)
- B the angle between the tangents drawn at A and B is neither 0 nor \(\frac{\pi}{2}\)
- C the tangent drawn at \(\mathrm{A}\) and the tangent drawn at \(\mathrm{B}\) are perpendicular to each other
- D the tangent drawn at \(\mathrm{A}\) is parallel to the tangent drawn at B
Answer & Solution
Correct Answer
(B) the angle between the tangents drawn at A and B is neither 0 nor \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
The given curve is : \(x=1+\frac{1}{y^2}\) ...(i) \(\Rightarrow 1=-2 \cdot \frac{1}{y^3} \cdot y^{\prime} \Rightarrow y^{\prime}=\frac{-y^2}{2}\) \(\Rightarrow \quad\left(y^{\prime}\right)_{(2,1)}=-\frac{1}{2}\). The equation of tangent at point \((2,1)\)…
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