Question

A tangent to the parabola ${{y}^{2}}=8x$ makes an angle of $45{}^\circ$ with the straight line y = 3x+5. Find its equation and point of contact.

Answer 1

A parabola is a curve where any point is at a fixed distance from a fixed point the focus and from a fixed line the directrix. ${{y}^{2}}=4ax$is the general form of a parabola with ‘a’ as the distance from the origin to the focus of the parabola and also the distance between the origin and directrix.
Equation of tangent the parabola ${{y}^{2}}=4ax$in slope form is $y=mx+\dfrac{a}{m}$ , where m is the slope of tangent.

Complete step by step answer:
Given equation of parabola is ${{y}^{2}}=8x$
Comparing ${{y}^{2}}=8x$ with ${{y}^{2}}=4ax$
4a=8, a=2
It is also given that the line y = 3x+5 makes an angle of $45{}^\circ$with the tangent of parabola
Comparing $y={{m}_{1}}x+c$, equation of straight line with slope ${{m}_{1}}$ and x-intercept c with y = 3x+5, we get that ${{m}_{1}}$=3 and c=5.
The slope of the line y=3x+5 is ${{m}_{1}}$=3 and the slope of the tangent to the parabola is m, angle between them is $45{}^\circ$.
The angle between the two lines with slopes ${{m}_{1}},{{m}_{2}}$ is $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ ,
We have $\theta =45{}^\circ$ and ${{m}_{1}}$=3,${{m}_{2}}=m$, substituting these values in $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
$\begin{aligned} & \Rightarrow \tan 45=\left| \dfrac{3-m}{1+3m} \right| \\\ & \Rightarrow 1=\left| \dfrac{3-m}{1+3m} \right| \\\ & \Rightarrow \dfrac{3-m}{1+3m}=\pm 1 \\\ & \Rightarrow \dfrac{3-m}{1+3m}=1,\dfrac{3-m}{1+3m}=-1 \\\ \end{aligned}$
Solving $\dfrac{3-m}{1+3m}$=1
$\begin{aligned} & \Rightarrow 3-m=1+3m \\\ & \Rightarrow 3-1=m+3m \\\ & \Rightarrow 2=4m \\\ & \Rightarrow m=\dfrac{1}{2} \\\ \end{aligned}$
Solving $\dfrac{3-m}{1+3m}=-1$
$\begin{aligned} & \Rightarrow 3-m=-1(1+3m) \\\ & \Rightarrow 3-m=-1-3m \\\ & \Rightarrow 3+1=m-3m \\\ & \Rightarrow 4=-2m \\\ & \Rightarrow m=-2 \\\ \end{aligned}$
The two values obtained for the slope of the tangent are $\dfrac{1}{2}$ and -2.
First let us solve considering the slope to be m=$\dfrac{1}{2}$
Substituting the value of slope which is m=$\dfrac{1}{2}$and ‘a’=2 in the equation of tangent $y=mx+\dfrac{a}{m}$
$\begin{aligned} & \Rightarrow y=\left( \dfrac{1}{2} \right)x+\dfrac{2}{\dfrac{1}{2}} \\\ & \Rightarrow y=\dfrac{x}{2}+4 \\\ & \Rightarrow 2y=x+8 \\\ & \Rightarrow x=2y-8 \\\ \end{aligned}$
The equation of tangent with slope $\dfrac{1}{2}$is x = 2y – 8.
To find the point of contact let us substitute x = 2y – 8 in the equation of parabola ${{y}^{2}}=8x$.
$\begin{aligned} & \Rightarrow {{y}^{2}}=8(2y-8) \\\ & \Rightarrow {{y}^{2}}=16y-64 \\\ & \Rightarrow {{y}^{2}}-16y+64=0 \\\ & \Rightarrow {{\left( y-8 \right)}^{2}}=0 \\\ & \Rightarrow y=8 \\\ \end{aligned}$
Substituting the value of y = 8 in x = 2y -8 we get
$\begin{aligned} & \Rightarrow x=2(8)-8 \\\ & \Rightarrow x=16-8 \\\ & \Rightarrow x=8 \\\ \end{aligned}$
The point of contact of the tangent x = 2y – 8 with the parabola ${{y}^{2}}=8x$is (8,8)
Now, let us solve by considering the slope to be m=-2
Substituting the value of slope which is m=$-2$and ‘a’=2 in the equation of tangent $y=mx+\dfrac{a}{m}$
$\begin{aligned} & \Rightarrow y=\left( -2 \right)x+\dfrac{2}{-2} \\\ & \Rightarrow y=-2x-1 \\\ & \Rightarrow y=-(2x+1) \\\ \end{aligned}$
The equation of tangent with slope $-2$ is y = - 2x - 1
To find the point of contact let us substitute y = -2x– 1 in the equation of parabola ${{y}^{2}}=8x$.
$\begin{aligned} & \Rightarrow {{\left( -2x-1 \right)}^{2}}=8x \\\ & \Rightarrow 4{{x}^{2}}+1+4x=8x \\\ & \Rightarrow 4{{x}^{2}}-4x+1=0 \\\ & \Rightarrow {{\left( 2x-1 \right)}^{2}}=0 \\\ & \Rightarrow x=\dfrac{1}{2} \\\ \end{aligned}$
Substituting the value of x = $\dfrac{1}{2}$ in y = - 2x - 1 we get
$\begin{aligned} & \Rightarrow y=-2(\dfrac{1}{2})-1 \\\ & \Rightarrow y=-1-1 \\\ & \Rightarrow y=-2 \\\ \end{aligned}$

The other point of contact of the tangent y = -2x– 1 with the parabola ${{y}^{2}}=8x$ is $\left( \dfrac{1}{2},-2 \right)$

Note: Here the equation of parabola is given in the general form so calculation was easier. The equation of tangent would be varying with the equation of type of parabola given. If the equation of tangent to the parabola${{y}^{2}}=4ax$ is $y=mx+\dfrac{a}{m}$, the point of contact would be $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$ . Direct substitution in the formula will also yield the same answer.