MHT CET · Maths · Differential Equations
The differential equation of all parabolas, whose axes are parallel to \(\mathrm{Y}\)-axis, is
- A \(y_3=1\)
- B \(y_3=-1\)
- C \(y_3=0\)
- D \(yy_3+y_1=0\)
Answer & Solution
Correct Answer
(C) \(y_3=0\)
Step-by-step Solution
Detailed explanation
Parabola whose axes are parallel to \(\mathrm{Y}\)-axis. Vertex is not \((0,0)\)
Equation becomes
\((x-\mathrm{h})^2=4 \mathrm{~b}(y-\mathrm{k})\)
Differentiating w.r.t. \(x\), we get
\(2(x-h)=4 b\left(\frac{d y}{d x}\right)\)
Again differentiating w.r.t. \(x\), we get
\(2=4 \mathrm{~b}\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)\)
Again differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& 0=4 b\left(\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}\right) \\
\therefore \quad & \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=0 \\
& \text { i.e., } y_3=0
\end{aligned}\)
Equation becomes
\((x-\mathrm{h})^2=4 \mathrm{~b}(y-\mathrm{k})\)
Differentiating w.r.t. \(x\), we get
\(2(x-h)=4 b\left(\frac{d y}{d x}\right)\)
Again differentiating w.r.t. \(x\), we get
\(2=4 \mathrm{~b}\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)\)
Again differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& 0=4 b\left(\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}\right) \\
\therefore \quad & \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=0 \\
& \text { i.e., } y_3=0
\end{aligned}\)
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