ExamBro
ExamBro
TS EAMCET · Maths · Application of Derivatives

\(\mathrm{P}(5,2)\) is a point on the curve \(y=f(x)\) and \(\frac{7}{2}\) is the slope of the tangent to the curve at P. The area of the triangle (in sq. units) formed by the tangent and the normal to the curve at P with \(x\)-axis is

  1. A \(35\)
  2. B \(\frac{35}{2}\)
  3. C \(\frac{53}{7}\)
  4. D \(\frac{53}{14}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{53}{7}\)

Step-by-step Solution

Detailed explanation

Tangent at \(P(5,2)\): \(y-2 = \frac{7}{2}(x-5) \Rightarrow 7x-2y-31=0\). X-intercept: \(x_T = \frac{31}{7}\). Normal at \(P(5,2)\): \(m_N = -\frac{2}{7} \Rightarrow y-2 = -\frac{2}{7}(x-5) \Rightarrow 2x+7y-24=0\). X-intercept: \(x_N = 12\). Base of triangle…
From TS EAMCET
Explore more questions on app