Question

Find the equation of the tangents to the ellipse \[2{{x}^{2}}+{{y}^{2}}=8\] which are
(i) parallel to \[x-2y=4\]
(ii) perpendicular to \[x+y+2=0\]

Answer 1

Hint: Differentiate the equation of the ellipse with respect to \[dx\] and get the value of \[\dfrac{dy}{dx}\] .
Compare the equation of the lines \[x-2y=4\] and \[x+y+2=0\] with the standard equation of the line, \[y=mx+c\] to obtain the slope of both lines. We know the property that the slope of two parallel lines is equal to each other. Also, the product of the slope of two perpendicular lines is equal to -1. Now, use these properties and get the slope and coordinates of points in both cases. We know the formula for the equation of a straight line having slope equal to \[m\] and passing through a point having coordinate \[\left( {{x}_{1}},{{y}_{1}} \right)\] , \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\] . Solve it further and get the equation of the tangents in both cases.

Complete answer:
According to the question, we are given that,
The equation of the ellipse = \[2{{x}^{2}}+{{y}^{2}}=8\] ………………………………………(1)
We know the property that the slope of a curve is equal to \[\dfrac{dy}{dx}\] ……………………………………………(2)
Now, differentiating the equation (1) with respect to \[dx\] , we get
 \[\Rightarrow \dfrac{d}{dx}\left( 2{{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 8 \right)\] ……………………………………………(3)
Using the chain rule and simplifying equation (9), we get
 \[\Rightarrow \dfrac{d}{dx}\left( 2{{x}^{2}} \right)+\dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}=0\] …………………………………………(4)
We know the formula, \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\] ……………………………………..(5)
Now, from equation (4) and equation (4), we get
\[\begin{align}
  & \Rightarrow 4x+2y\times \dfrac{dy}{dx}=0 \\
 & \Rightarrow 2y\times \dfrac{dy}{dx}=-4x \\
\end{align}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-2x}{y}\] …………………………………………(6)
From equation (2) and equation (6), we get
The slope of the ellipse = \[\dfrac{-2x}{y}\] ………………………………………..(7)
We have to find the equation of the tangent parallel to the line \[x-2y=4\] …………………………….(8)
Modifying equation (8), we get
\[\begin{align}
  & \Rightarrow x-2y=4 \\
 & \Rightarrow x-4=2y \\
 & \Rightarrow \dfrac{x}{2}-2=y \\
\end{align}\]
\[\Rightarrow y=\dfrac{x}{2}-2\] ……………………………………(9)
We know the standard equation of the straight line, \[y=mx+c\] where \[m\] is the slope of the straight line ………………………………………………(10)
Now, on comparing equation (9) and equation (10), we get
The slope of the line = \[\dfrac{1}{2}\] ……………………………………………(11)
We also know the property that the slope of parallel lines are equal ………………………….….(12)
Now, from equation (7) and equation (12), we get
The slope of the tangent to the ellipse = \[\dfrac{1}{2}\]
\[\Rightarrow \dfrac{-2x}{y}=\dfrac{1}{2}\]
\[\Rightarrow -4x=y\] …………………………………….(13)
Now, from equation (1) and equation (13), we get
 \[\begin{align}
  & \Rightarrow 2{{x}^{2}}+{{\left( -4x \right)}^{2}}=8 \\
 & \Rightarrow 2{{x}^{2}}+16{{x}^{2}}=8 \\
 & \Rightarrow 18{{x}^{2}}=8 \\
 & \Rightarrow {{x}^{2}}=\dfrac{8}{18} \\
 & \Rightarrow {{x}^{2}}=\dfrac{4}{9} \\
\end{align}\]
\[\Rightarrow x=\dfrac{2}{3}\] ………………………………………..(14)
Putting \[x=\dfrac{2}{3}\] in equation (13), we get
\[\Rightarrow -4\times \dfrac{2}{3}=y\]
\[\Rightarrow \dfrac{-8}{3}=y\] …………………………………………..(15)
From equation (14) and equation (15), we have got the x-coordinate and y-coordinate.
We know the formula for the equation of a straight line having slope equal to \[m\] and passing through a point having coordinate \[\left( {{x}_{1}},{{y}_{1}} \right)\] , \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\] ………………………………………(16)
Now, from equation (14) and equation (15), and using the formula shown in equation (16), we get
\[\begin{align}
  & \Rightarrow \left( y-\left( \dfrac{-8}{3} \right) \right)=\dfrac{1}{2}\left( x-\dfrac{2}{3} \right) \\
 & \Rightarrow y+\dfrac{8}{3}=\dfrac{x}{2}-\dfrac{1}{3} \\
 & \Rightarrow y=\dfrac{x}{2}-\dfrac{9}{3} \\
\end{align}\]
\[\Rightarrow 2y=x-6\] ………………………………………..(17)
We also have to find the equation of the tangent perpendicular to the line \[x+y+2=0\] ……………………………. (18)
Modifying equation (18), we get
\[\Rightarrow x+y+2=0\]
\[\Rightarrow y=-x-2\] ……………………………………(19)
Now, on comparing equation (10) and equation (19), we get
The slope of the line = \[-1\] ……………………………………………(20)
We also know the property that the product of slope two perpendicular lines is equal to -1.
Since the tangent is perpendicular to the line \[y+x+2\] so, the product of the slope of the tangent and the line is equal to -1.
\[\Rightarrow \] The slope of the tangent \[\times \] The slope of the line = -1 ……………………………………..(21)
From equation (20) and equation (21), we get
 \[\Rightarrow \] The slope of the tangent \[\times \] -1 = -1
\[\Rightarrow \] The slope of the tangent = 1 ……………………………………………..(22)
Now, from equation (7) and equation (22), we get
\[\Rightarrow \dfrac{-2x}{y}=1\]
\[\Rightarrow -2x=y\] …………………………………….(23)
Now, from equation (1) and equation (23), we get
 \[\begin{align}
  & \Rightarrow 2{{x}^{2}}+{{\left( -2x \right)}^{2}}=8 \\
 & \Rightarrow 2{{x}^{2}}+4{{x}^{2}}=8 \\
 & \Rightarrow 6{{x}^{2}}=8 \\
 & \Rightarrow {{x}^{2}}=\dfrac{8}{6} \\
\end{align}\]
\[\Rightarrow x=\dfrac{2}{\sqrt{3}}\] ………………………………………..(24)
Putting \[x=\dfrac{2}{\sqrt{3}}\] in equation (23), we get
\[\Rightarrow -2\times \dfrac{2}{\sqrt{3}}=y\]
\[\Rightarrow \dfrac{-4}{\sqrt{3}}=y\] …………………………………………..(25)
From equation (24) and equation (25), we have got the x-coordinate and y-coordinate.
We know the formula for the equation of a straight line having slope equal to \[m\] and passing through a point having coordinate \[\left( {{x}_{1}},{{y}_{1}} \right)\] , \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\] ………………………………………(26)
Now, from equation (22), equation (24), and equation (25), and using the formula shown in equation (16), we get
\[\begin{align}
  & \Rightarrow \left( y-\left( \dfrac{-4}{\sqrt{3}} \right) \right)=1\left( x-\dfrac{2}{\sqrt{3}} \right) \\
 & \Rightarrow y+\dfrac{4}{\sqrt{3}}=x-\dfrac{2}{\sqrt{3}} \\
 & \Rightarrow y=x-\dfrac{6}{\sqrt{3}} \\
\end{align}\]
\[\Rightarrow y=x-2\sqrt{3}\] ………………………………………..(27)
From equation (17) and equation (27), we have the equation of the tangents.
Therefore, the equation of the tangents parallel to the line \[x-2y=4\] and perpendicular to \[x+y+2=0\] are \[y=-x-2\] and \[y=x-4\sqrt{3}\] respectively.

Note:
For this question, we need to remember two points. The first point is, the slope of two parallel lines are equal to each other. The second point is, the product of the slope of two perpendicular lines is equal to -1.