JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(C\) be a curve given by \(y\left( x \right) = 1 + \sqrt {4x - 3} ,x > \frac{3}{4}\).If \(P\) is a point on \(C\), such that the tangent at \(P\) has slope \(\frac{2}{3}\) , then a point through which the normal at \(P\) passes, is
- A \((1, 7)\)
- B \((3, -4)\)
- C \((4, -3)\)
- D \((2, 3)\)
Answer & Solution
Correct Answer
(A) \((1, 7)\)
Step-by-step Solution
Detailed explanation
\(\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {4x - 3} }} \times 4 = \frac{2}{3}\) \( \Rightarrow 4x - 3 = 9\) \( \Rightarrow x = 3\) so, \(y=4\) Equation of normal at \(P(3,4)\) is \(y - 4 = - \frac{3}{2}\left( {x - 3} \right)\) i.e. \(2y - 8 = - 3x + 9\)…
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