
The differential equation of all parabolas with axis parallel to axis of $y-axis$ is:-
A) ${{y}_{2}}=2{{y}_{1}}+x$
b.) ${{y}_{3}}=2{{y}_{1}}$
c.) \[y_{2}^{3}={{y}_{1}}\]
d.) ${{y}_{3}}=0$
Answer
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Hint: In order to solve this particular question, we need to differentiate the standard equation of the parabola, and then again we further differentiate the equation we get. In order to get our final result, we will differentiate the equation of the parabola, thrice in this particular question.
Complete step by step solution: The general equation of parabola parallel to $y-axis$ and having center $\left( h,\text{k} \right)$ and having the distance ‘a’ from vertex to the focus is:- ${{\left( x-h \right)}^{2}}=4a\ (y-k).$
${{\left( x-h \right)}^{2}}=4a\ (y-k).(\text{i})$
We need to differentiate this equation further, differentiating the equation (i) once, we get;
$=\dfrac{dy}{dx}{{\left( x-h \right)}^{2}}=\dfrac{dy}{dx}4a\left( y-\text{k} \right)$
$\Rightarrow 2\left( x-h \right)=4a\dfrac{dy}{dx}(\text{ii})$
[ as because 'k' is constant here].
Now again on differentiating equation (ii) we get;
$\Rightarrow \dfrac{d}{dx}\left( 2\left( x-h \right) \right)=\dfrac{d}{dx}\left( 4a\dfrac{dy}{dx} \right)$
\[\Rightarrow 2=4a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2}{2a}(\text{iii})$
Differentiating equation (iii) once again ; on both the sides, we get;
$\Rightarrow \dfrac{dy}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\ \dfrac{dy}{dx}\left( \dfrac{1}{2a} \right)$
$\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=0$
$\Rightarrow{y}_3=0 $
Therefore, option (d) is the correct answer to this question.
Note: A parabola is symmetric with its axis. If the equation has a $y^2$ term, then the axis of symmetry is along the x-axis and if the equation has an $x^2$ term, then the axis of symmetry is along the y-axis.
For axis of symmetry along the x-axis:
It opens to the right if the coefficient of x is positive.
It opens to the left if the coefficient of x is negative.
For axis of symmetry along the y-axis:
It opens upwards if the coefficient of y is positive.
It opens downwards if the coefficient of y is negative.
The standard equation of the parabola is:-
${{\left( x-h \right)}^{2}}=4a\left( y-\text{k} \right);$
where, $\left( h,\text{k} \right)$ is the center of the parabola.
Complete step by step solution: The general equation of parabola parallel to $y-axis$ and having center $\left( h,\text{k} \right)$ and having the distance ‘a’ from vertex to the focus is:- ${{\left( x-h \right)}^{2}}=4a\ (y-k).$
${{\left( x-h \right)}^{2}}=4a\ (y-k).(\text{i})$
We need to differentiate this equation further, differentiating the equation (i) once, we get;
$=\dfrac{dy}{dx}{{\left( x-h \right)}^{2}}=\dfrac{dy}{dx}4a\left( y-\text{k} \right)$
$\Rightarrow 2\left( x-h \right)=4a\dfrac{dy}{dx}(\text{ii})$
[ as because 'k' is constant here].
Now again on differentiating equation (ii) we get;
$\Rightarrow \dfrac{d}{dx}\left( 2\left( x-h \right) \right)=\dfrac{d}{dx}\left( 4a\dfrac{dy}{dx} \right)$
\[\Rightarrow 2=4a\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2}{2a}(\text{iii})$
Differentiating equation (iii) once again ; on both the sides, we get;
$\Rightarrow \dfrac{dy}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\ \dfrac{dy}{dx}\left( \dfrac{1}{2a} \right)$
$\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=0$
$\Rightarrow{y}_3=0 $
Therefore, option (d) is the correct answer to this question.
Note: A parabola is symmetric with its axis. If the equation has a $y^2$ term, then the axis of symmetry is along the x-axis and if the equation has an $x^2$ term, then the axis of symmetry is along the y-axis.
For axis of symmetry along the x-axis:
It opens to the right if the coefficient of x is positive.
It opens to the left if the coefficient of x is negative.
For axis of symmetry along the y-axis:
It opens upwards if the coefficient of y is positive.
It opens downwards if the coefficient of y is negative.
The standard equation of the parabola is:-
${{\left( x-h \right)}^{2}}=4a\left( y-\text{k} \right);$
where, $\left( h,\text{k} \right)$ is the center of the parabola.
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