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Question: A tangent to the parabola \({y^2} = 8x\) makes an angle of \({45^ \circ }\) with the straight line \...

A tangent to the parabola y2=8x{y^2} = 8x makes an angle of 45∘{45^ \circ } with the straight line y=3x+5y = 3x + 5 Then points of contact are.

Explanation

Solution

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 ⇒tan⁡θ=∣m1−m21+m1m2∣ \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 θ=45\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 y2=8x{y^2} = 8x and an equation of a straight line as y=3x+5y = 3x + 5.
According to the question the tangent of the parabola and the straight line given makes an angle of 45∘{45^ \circ }.
Let’s first find out the slope of the given straight liney=3x+5y = 3x + 5 by comparing it to the slope-intercept form y=m1x+cy = {m_1}x + c, where m1{m_1} is the slope.
We get the slope m1=3{m_1} = 3.
Now let the slope of the tangent to the parabola be m2{m_2}
Let's find out the value of m2{m_2} by using the formula of angle between two straight line as ⇒tan⁡θ=∣m1−m21+m1m2∣ \Rightarrow \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|
Now putting the value of θ=45∘\theta = {45^ \circ }and m1{m_1}, in above formula and solving it for m2{m_2}
⇒tan⁡45∘=∣3−m21+(3)m2∣\Rightarrow \tan {45^ \circ } = \left| {\dfrac{{3 - {m_2}}}{{1 + \left( 3 \right){m_2}}}} \right|
The value of tan⁡45∘=1\tan {45^ \circ } = 1
⇒1=∣3−m21+(3)m2∣\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 form2{m_2}
⇒1=3−m21+(3)m2,−1=3−m21+(3)m2 ⇒1(1+3m2)=3−m2,−1(1+3m2)=3−m2 ⇒1+3m2=3−m2,−1−3m2=3−m2 ⇒4m2=2,−2m2=4  \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 \\\
⇒m2=12,m2=−2\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 y2=8x{y^2} = 8x with respect to xx
2ydydx=8 dydx=82y dydx=4y y′=4y  2y\dfrac{{dy}}{{dx}} = 8 \\\ \dfrac{{dy}}{{dx}} = \dfrac{8}{{2y}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{4}{y} \\\ y' = \dfrac{4}{y} \\\
y′y'is nothing but the slope that we have obtained in equation (1).
1.If m2=y′=12{m_2} = y' = \dfrac{1}{2}
12=4y y=8  \dfrac{1}{2} = \dfrac{4}{y} \\\ y = 8 \\\
Putting y=8y = 8 in the equation of parabola . we get
(8)2=8x x=8  {\left( 8 \right)^2} = 8x \\\ x = 8 \\\
Required point (8,8)\left( {8,8} \right)
2.If m2=y′=−2{m_2} = y' = - 2
\-2=4y y=−2  \- 2 = \dfrac{4}{y} \\\ y = - 2 \\\
Putting y=−2y = - 2 in the equation of parabola . we get
(−2)2=8x x=48=12  {\left( { - 2} \right)^2} = 8x \\\ x = \dfrac{4}{8} = \dfrac{1}{2} \\\
Required point (12,−2)\left( {\dfrac{1}{2}, - 2} \right)

Therefore, the point of contacts are (12,−2)\left( {\dfrac{1}{2}, - 2} \right) and (8,8)\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.