Question

A tangent to the parabola ${y^2} = 8x$ makes an angle of ${45^ \circ }$ with the straight line $y = 3x + 5$ Then points of contact are.

Answer 1

Hint: In order to determine the point of contacts for the given parabola and straight line, find out the slope of the parabola by using the formula of angle between two lines as $ \Rightarrow \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$. By putting the slope of straight line which can be obtained by comparing it with the slope-intercept form and $\theta = 45$.now find the derivative of parabola with respect to x and remember the derivative is nothing but the slope, so put the slope obtained one by one to obtain the value of y . use this value of y to obtain the respective x using the equation of parabola. The pair of values of x and y are the required points of contact.

Complete step by step answer:
We are given an equation of parabola as ${y^2} = 8x$ and an equation of a straight line as $y = 3x + 5$.
According to the question the tangent of the parabola and the straight line given makes an angle of ${45^ \circ }$.
Let’s first find out the slope of the given straight line$y = 3x + 5$ by comparing it to the slope-intercept form $y = {m_1}x + c$, where ${m_1}$ is the slope.
We get the slope ${m_1} = 3$.
Now let the slope of the tangent to the parabola be ${m_2}$
Let's find out the value of ${m_2}$ by using the formula of angle between two straight line as $ \Rightarrow \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$
Now putting the value of $\theta = {45^ \circ }$and ${m_1}$, in above formula and solving it for ${m_2}$
$ \Rightarrow \tan {45^ \circ } = \left| {\dfrac{{3 - {m_2}}}{{1 + \left( 3 \right){m_2}}}} \right|$
The value of $\tan {45^ \circ } = 1$
$ \Rightarrow 1 = \left| {\dfrac{{3 - {m_2}}}{{1 + \left( 3 \right){m_2}}}} \right|$
Remember Whenever the modulo is removed it results in $ \pm $. We will get two values for${m_2}$
$
   \Rightarrow 1 = \dfrac{{3 - {m_2}}}{{1 + \left( 3 \right){m_2}}}, - 1 = \dfrac{{3 - {m_2}}}{{1 + \left( 3 \right){m_2}}} \\
   \Rightarrow 1\left( {1 + 3{m_2}} \right) = 3 - {m_2}, - 1\left( {1 + 3{m_2}} \right) = 3 - {m_2} \\
   \Rightarrow 1 + 3{m_2} = 3 - {m_2}, - 1 - 3{m_2} = 3 - {m_2} \\
   \Rightarrow 4{m_2} = 2, - 2{m_2} = 4 \\
 $
$ \Rightarrow {m_2} = \dfrac{1}{2},{m_2} = - 2$-(1)
Hence,We have obtained the value for slope of tangent to parabola .
Let’s now find the tangent by differentiating the equation of parabola ${y^2} = 8x$ with respect to $x$
$
  2y\dfrac{{dy}}{{dx}} = 8 \\
  \dfrac{{dy}}{{dx}} = \dfrac{8}{{2y}} \\
  \dfrac{{dy}}{{dx}} = \dfrac{4}{y} \\
  y' = \dfrac{4}{y} \\
 $
$y'$is nothing but the slope that we have obtained in equation (1).
1.If ${m_2} = y' = \dfrac{1}{2}$
$
  \dfrac{1}{2} = \dfrac{4}{y} \\
  y = 8 \\
 $
Putting $y = 8$ in the equation of parabola . we get
$
  {\left( 8 \right)^2} = 8x \\
  x = 8 \\
 $
Required point $\left( {8,8} \right)$
2.If ${m_2} = y' = - 2$
$
   - 2 = \dfrac{4}{y} \\
  y = - 2 \\
 $
Putting $y = - 2$ in the equation of parabola . we get
$
  {\left( { - 2} \right)^2} = 8x \\
  x = \dfrac{4}{8} = \dfrac{1}{2} \\
 $
Required point $\left( {\dfrac{1}{2}, - 2} \right)$

Therefore, the point of contacts are $\left( {\dfrac{1}{2}, - 2} \right)$ and $\left( {8,8} \right)$

Note: 1. Remember whenever there is a modulus , the solution will contain the both positive and negative solution for the given expression.
2.Don’t forget to cross check your answer.