WBJEE · Maths · Application of Derivatives
Tangent is drawn at any point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) on a curve, which passes through \((1,1)\). The tangent cuts X-axis and Y-axis at A and B respectively. If AP:BP = 3:1, then
- A the differential equation of the curve is \(3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0\)
- B the differential equation of the curve is \(3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0\)
- C the curve passes through \(\left(\frac{1}{8}, 2\right)\)
- D the normal at \((1,1)\) is \(x+3 y=4\)
Answer & Solution
Correct Answer
(C) the curve passes through \(\left(\frac{1}{8}, 2\right)\)
Step-by-step Solution
Detailed explanation
Hint: \(\frac{P A}{P B}=\frac{3}{1}\) Equation of tangent \(\mathrm{AB}\) is \(\mathrm{Y}-\mathrm{y}=\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}(\mathrm{X}-\mathrm{x})\) \(\mathrm{A}\left(\frac{\mathrm{xy}^{\prime}-\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\) and…
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