Answer
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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 \[\dfrac{-dx}{dy}\].
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