A parabola pases through (1,2) and satisfies the differential equation (d y)/(d x)=(2 y)/(x), x > 0, y > 0 . What is the directrix of the parabola?

Correct answer: A. y=-(1)/(8) aligned & Since, (d y)/(d x)=(2 y)/(x) & ∫ (d y)/(y)=∫ (2 d x)/(x) y=2 x+ c y=x^2 c & it satisfy (1,2) & 2=1 C C=2 y=2 x^2 aligned Since, y=2 x^2 x^2=(1)/(2) y x^2=4 · (1)/(8) y So, directrix of Parabola is y=-(1)/(8)

NDA 2023

Exam: NDA
Year: 2023
Subject: Mathematics
Chapter: Differential Equations

NDA 2023 · Mathematics · Differential Equations

A parabola pases through (1,2) and satisfies the differential...

A parabola pases through

(1,2)

and satisfies the differential equation

\frac{d y}{d x}=\frac{2 y}{x}, x > 0, y > 0

. What is the directrix of the parabola?

Options

A $y=-\frac{1}{8}$
B $y=\frac{1}{8}$
C $x=-\frac{1}{8}$
D $x=\frac{1}{8}$

Correct answer

A. $y=-\frac{1}{8}$

Step-by-step solution

\begin{aligned} & \text{ Since, } \frac{d y}{d x}=\frac{2 y}{x} \\ & \int \frac{d y}{y}=\int \frac{2 d x}{x} \Rightarrow \ln y=2 \ln x+\ln c \Rightarrow y=x^2 c \\ & \text{it satisfy }(1,2) \\ & \quad \Rightarrow 2=1 C \Rightarrow C=2 \Rightarrow y=2 x^2\end{aligned}

Since,

y=2 x^2 \Rightarrow x^2=\frac{1}{2} y \Rightarrow x^2=4 \cdot \frac{1}{8} y

So, directrix of Parabola is

y=-\frac{1}{8}

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