AP EAMCET · Maths · Application of Derivatives
If the slope of the tangent at a point \((\mathrm{x}, \mathrm{y})\) on a curve is \(\frac{y-4}{x-3}\) and the curve passes through \((4,3)\), then the point where it cuts the line \(\mathrm{y}=\mathrm{x}\) is
- A \((1,1)\)
- B \((3,3)\)
- C \(\left(\frac{7}{2}, \frac{7}{2}\right)\)
- D \(\left(-\frac{5}{2},-\frac{5}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{7}{2}, \frac{7}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{y-4}{x-3}\) \(\Rightarrow \int \frac{d x}{x-3}=\int \frac{d y}{y-4}\) \(\Rightarrow \ln (x-3 \mid)=\ln (|y-4|)+c\) as curve passes through \((4,3)\) \(\Rightarrow \ln 1=\ln 1+c\) \(\Rightarrow \quad c=0\) \(\Rightarrow \ln |x-3|-\ln |y-4|=0\)…
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