Question

Statement 1: The slope of the tangent at any point \[P\]on a parabola, whose axis is the axis of \[X\] and vertex is at the origin, is inversely proportional to the ordinate of the point \[P\].
Statement 2: The system of parabolas \[{{Y}^{2}}=4ax\] satisfies a differential equation of degree 1 and order 1.
(1) Statement 1 is true Statement 2 is true, Statement 2is a correct explanation for Statement 1.
(2) Statement 1 is true and Statement 2 is false.
(3) Statement 1 is true Statement 2 is true, Statement 2is not a correct explanation for Statement 1.
(4) Statement 1 is false and Statement 2 is true.

Answer 1

Hint: Consider a standard parabola equation and find the slope using the differentiation method. Then, find out the differential equation by eliminating the constant term.

Complete step by step answer:
Let us consider the parabola as \[{{Y}^{2}}=4ax\] and assume a variable point P on it as \[\left( a{{t}^{2}},2at \right)\].
Now differentiating the parabola equation, we have:
\[\begin{align}
  & \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( 4ax \right)}{dx} \\
 & 2y.\dfrac{dy}{dx}=4a \\
 & \dfrac{dy}{dx}=\dfrac{2a}{y}.......(1) \\
 & \\
\end{align}\]
As \[\dfrac{dy}{dx}\] refers to the slope of the tangent of any given curve, let us find the slope of tangent at our required point P\[\left( a{{t}^{2}},2at \right)\].
\[\begin{align}
  & {{\left( \dfrac{dy}{dx} \right)}_{\left( a{{t}^{2}},2at \right)}}=\dfrac{2a}{2at} \\
 & {{\left( \dfrac{dy}{dx} \right)}_{\left( a{{t}^{2}},2at \right)}}=\dfrac{1}{t} \\
\end{align}\]
From the above equation we can say that the slope of tangent at point \[P\left( a{{t}^{2}},2at \right)\] is inversely proportional to “t”, that is, to the y coordinate of point P.
So, it can be said that statement 1 is true.

For statement 2:
From equation (1) we have that,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{2a}{y} \\
 & a=\dfrac{y}{2}.\dfrac{dy}{dx} \\
\end{align}\]
Substituting \[a=\dfrac{y}{2}.\dfrac{dy}{dx}\] in \[{{Y}^{2}}=4ax\], to find out the required differential equation, we have:
\[\begin{align}
  & {{y}^{2}}=4x.\dfrac{y}{2}\dfrac{dy}{dx} \\
 & {{Y}^{2}}=2xy\dfrac{dy}{dx} \\
 & {{Y}^{2}}=2x.\dfrac{dy}{dx} \\
\end{align}\]
The order of the above differential equation is 1 and the degree is also 1. Since, the order of a differential equation is the highest numbered derivative and degree is the highest power to which the derivative is raised.
Therefore, Both the statements are true, but statement 2 is not the correct explanation of statement 2, as statement 2 does not explain about statement 1.

Hence, option 2 is the right answer.

Note: We must be aware that the order of a differential equation is the highest numbered derivative and degree is the highest power to which the derivative is raised.