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:

& \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{aligned}$$ As $$\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{aligned} & {{\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{aligned}$$ From the above equation we can say that the slope of tangent at point $$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{aligned} & \dfrac{dy}{dx}=\dfrac{2a}{y} \\\ & a=\dfrac{y}{2}.\dfrac{dy}{dx} \\\ \end{aligned}$$ Substituting $$a=\dfrac{y}{2}.\dfrac{dy}{dx}$$ in $${{Y}^{2}}=4ax$$, to find out the required differential equation, we have: $$\begin{aligned} & {{y}^{2}}=4x.\dfrac{y}{2}\dfrac{dy}{dx} \\\ & {{Y}^{2}}=2xy\dfrac{dy}{dx} \\\ & {{Y}^{2}}=2x.\dfrac{dy}{dx} \\\ \end{aligned}$$ 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.