Question

Let C be a curve $y(x) = 1 + \sqrt {4x - 3} $, $x > \dfrac{3}{4}$. If P is a point on C such that the tangent at P has slope $\dfrac{2}{3}$, then a point through which the normal at P passes is:
$
  a) \,(1,7) \\
  b) \,(3, - 4) \\
  c) \,(4, - 3) \\
  d) \,(2,3) \\
 $

Answer 1

Hint: Slope of tangent of any equation in term of $y = f(x)$ is given by $\dfrac{{dy}}{{dx}}$ and if ${m_1}$ is the slope of tangent and ${m_2}$ is the slope of normal then their product will be equal to $ - 1$ i.e. ${m_1} \times {m_2} = - 1$.

Complete step-by-step answer:
So here a curve is given by equation $y(x) = 1 + \sqrt {4x - 3} ,\,x > \dfrac{3}{4}$
Its tangent at point P has slope $\dfrac{2}{3}$.
We need to find the point through with normal at P passes.
So first of all we know that slope of tangent is given by $\dfrac{{dy}}{{dx}}$
And let us assume the point P has coordinates $\left( {\alpha ,\beta } \right)$
So let us find $\dfrac{{dy}}{{dx}}$at point $\left( {\alpha ,\beta } \right)$
Given $y(x) = 1 + \sqrt {4x - 3} $
Now differentiating with respect to $x$both sides
$
  \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{4}{{\sqrt {4x - 3} }} \\
  \dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {4x - 3} }} \\
 $
So at point P$\left( {\alpha ,\beta } \right)$, it is given that slope is $\dfrac{2}{3}$. So, at $x = \alpha \,\,\,\& \,\,\,y = \beta \,\,\,\,\,\,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{2}{3}$
And $\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {4x - 3} }}$
Putting $x = \alpha \,\,\,\& \,\,\,y = \beta \,$we get
$
  \dfrac{2}{{\sqrt {4\alpha - 3} }} = \dfrac{2}{3} \\
  \sqrt {4\alpha - 3} = 3 \\
 $
Squaring both sides,
$
  4\alpha - 3 = {3^2} \\
  4\alpha - 3 = 9 \\
  4\alpha = 12 \\
  \alpha = 3 \\
 $
Then we know that $y(x) = 1 + \sqrt {4x - 3} $. So
$
  \beta = \;1 + \sqrt {4\alpha - 3} \\
  \beta = \;1 + \sqrt {4\left( 3 \right) - 3} \\
  \beta = \;1 + \sqrt 9 \\
  \beta = \;1 + 3 \\
  \beta = \;4 \\
 $
We get that coordinates of point P are P$\left( {3,4} \right)$
Now we know that slope of tangent is $\dfrac{2}{3}$and ${m_{\tan gent}} \times {m_{normal}} = - 1$as both are perpendicular.
So,
 $
  \dfrac{2}{3} \times {m_{normal}} = - 1 \\
  {m_{normal}} = - \dfrac{3}{2} \\
 $
Now equation of normal can be written as
$y = {m_{normal}}x + c$and at point P it must satisfy P$\left( {3,4} \right)$. So,
$
  4 = - \dfrac{3}{2}\left( 3 \right) + c \\
  c = 4 + \dfrac{9}{2} \\
  c = \dfrac{{17}}{2} \\
 $
So equation of normal will become
$
  y = - \dfrac{3}{2}x + \dfrac{{17}}{2} \\
  2y + 3x = 17 \\
 $
Now let us check the option which satisfies this equation
Option A: $\left( {1,7} \right) \Rightarrow 14 + 3 = 17$.
So option A is correct.

Note: If two lines are given as $y = {m_1}x + c$ and $y = {m_2}x + c$ then angle between them is given by $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$. So as normal and tangent are at right angle to each other so angle is $90$ and $\tan \theta = \infty $therefore ${m_1}{m_2} + 1 = 0$ i.e. $({m_1}{m_2} = - 1)$