
If $2x+y+\lambda =0$ is the normal to the parabola \[{{y}^{2}}=8x\], then $\lambda =$
(a) -24
(b) 8
(c) -16
(d) 24
Answer
585k+ views
Hint: At first differentiate the parabola to make the subject $\left( \dfrac{dy}{dx} \right)$, then find the slope. Next find $\left( \dfrac{-dx}{dy} \right)$ which is slope of normal then compare it with given equation to find the point y then substitute in the equation \[{{y}^{2}}=8x\], to get the value of ‘x’. Then put the values of ‘x’ and ‘y’ in the equation $2x+y+\lambda =0$, to get the value of $\lambda $.
Complete step-by-step answer:
In the question we are given the equation of parabola \[{{y}^{2}}=8x\].
At first we will find the slope of parabola.
For finding the tangent we will use the process as shown below,
If the equation of parabola is \[{{y}^{2}}=4ax\] then the slope of the parabola will be found by just differentiating \[{{y}^{2}}=4ax\] with respect to ‘x’.
Therefore, for the equation of parabola asked in the question \[{{y}^{2}}=8x\].
The slope will be:
\[{{y}^{2}}=8x\]
After differentiating \[{{y}^{2}}=8x\] we get,
\[2y\dfrac{dy}{dx}=8\]
By using formula,
\[\dfrac{d}{dx}\left( {{y}^{2}} \right)=2y\dfrac{dy}{dx}\] and $\dfrac{d}{dx}\left( x \right)=1$
So, \[\]
$\dfrac{dy}{dx}=\dfrac{4}{y}$ at point (x,y).
The slope of normal at point (x, y) will be \[\dfrac{-dx}{dy}\] which is equal to $\left( \dfrac{-1}{\dfrac{dy}{dx}} \right)$
So, the slope of normal is $\dfrac{-1}{\dfrac{4}{y}}=\dfrac{-y}{4}.$
As, we know the equation of normal which is given by 2x + y + 7 = 0.
After rearranging the line we can write it as,
y = -2x – 7
The slope will be ‘-2’ as we know that if the line is y = mx + c then its slope is always ‘m’.
Now we know that the slope of normal is $\dfrac{-y}{4}$ and also -2. So we can write as,
$\dfrac{-y}{4}=-2$
So, y = 8
Now substituting y = 8 in equation ${{y}^{2}}=8x$ we will get the value of x.
So, ${{8}^{2}}=8x$
Then x = 8.
So, the point is (8, 8).
Now substituting (8, 8) in line $2x+y+\lambda =0$ to get the value of $\lambda $.
This can be simplified as,
$\begin{align}
& 2(8)+8+\lambda =0 \\
& \Rightarrow \lambda =-16-8=-24 \\
\end{align}$
So, the answer is option (a).
Note: Students before solving the problem should know the basics of differentiation and how to find the slope of tangent and normal. They should be also careful about the calculations so as to avoid errors.
Students generally get confused when finding the slope of parabola. They generally take the slope of parabola as $-\dfrac{dy}{dx}$ instead of \[\dfrac{-dx}{dy}\].
Complete step-by-step answer:
In the question we are given the equation of parabola \[{{y}^{2}}=8x\].
At first we will find the slope of parabola.
For finding the tangent we will use the process as shown below,
If the equation of parabola is \[{{y}^{2}}=4ax\] then the slope of the parabola will be found by just differentiating \[{{y}^{2}}=4ax\] with respect to ‘x’.
Therefore, for the equation of parabola asked in the question \[{{y}^{2}}=8x\].
The slope will be:
\[{{y}^{2}}=8x\]
After differentiating \[{{y}^{2}}=8x\] we get,
\[2y\dfrac{dy}{dx}=8\]
By using formula,
\[\dfrac{d}{dx}\left( {{y}^{2}} \right)=2y\dfrac{dy}{dx}\] and $\dfrac{d}{dx}\left( x \right)=1$
So, \[\]
$\dfrac{dy}{dx}=\dfrac{4}{y}$ at point (x,y).
The slope of normal at point (x, y) will be \[\dfrac{-dx}{dy}\] which is equal to $\left( \dfrac{-1}{\dfrac{dy}{dx}} \right)$
So, the slope of normal is $\dfrac{-1}{\dfrac{4}{y}}=\dfrac{-y}{4}.$
As, we know the equation of normal which is given by 2x + y + 7 = 0.
After rearranging the line we can write it as,
y = -2x – 7
The slope will be ‘-2’ as we know that if the line is y = mx + c then its slope is always ‘m’.
Now we know that the slope of normal is $\dfrac{-y}{4}$ and also -2. So we can write as,
$\dfrac{-y}{4}=-2$
So, y = 8
Now substituting y = 8 in equation ${{y}^{2}}=8x$ we will get the value of x.
So, ${{8}^{2}}=8x$
Then x = 8.
So, the point is (8, 8).
Now substituting (8, 8) in line $2x+y+\lambda =0$ to get the value of $\lambda $.
This can be simplified as,
$\begin{align}
& 2(8)+8+\lambda =0 \\
& \Rightarrow \lambda =-16-8=-24 \\
\end{align}$
So, the answer is option (a).
Note: Students before solving the problem should know the basics of differentiation and how to find the slope of tangent and normal. They should be also careful about the calculations so as to avoid errors.
Students generally get confused when finding the slope of parabola. They generally take the slope of parabola as $-\dfrac{dy}{dx}$ instead of \[\dfrac{-dx}{dy}\].
Recently Updated Pages
The height of a solid metal cylinder is 20cm Its r-class-10-maths-ICSE

If a train crossed a pole at a speed of 60kmhr in 30 class 10 physics CBSE

Name the Writs that the High Courts are empowered to class 10 social science CBSE

A tower is 5sqrt 3 meter high Find the angle of el-class-10-maths-CBSE

Immediate cause of variations of A Mutations B Environmental class 10 biology CBSE

A rectangular container whose base is a square of side class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Why is Sardar Vallabhbhai Patel called the Iron man class 10 social science CBSE

Tropical deciduous trees shed their leaves in the dry class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Write a letter to the principal requesting him to grant class 10 english CBSE

Leap year has days A 365 B 366 C 367 D 368 class 10 maths CBSE

