MHT CET · Maths · Differential Equations
The equation of the curve passing through the point \((0,-2)\) given that at any point \((x, y)\) on the curve, the product of the slope of its tangent and \(y\)-co-ordinate of the point is equal to the \(x\)-co-ordinate of the point, is
- A \(y^2+x^2=4\)
- B \(y^2-x^2=4\)
- C \(2 y^2+x^2=8\)
- D \(4 y^2+3 x^2=16\)
Answer & Solution
Correct Answer
(B) \(y^2-x^2=4\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& A / Q \frac{d y}{d x} \cdot y=x \\
& \Rightarrow \int y d y=\int x d x \\
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c
\end{aligned}\)
putting \(x=0\) and \(y=-2\) we get \(c=2\)
\(\begin{aligned}
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+2 \\
& \Rightarrow y^2-x^2=4
\end{aligned}\)
& A / Q \frac{d y}{d x} \cdot y=x \\
& \Rightarrow \int y d y=\int x d x \\
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c
\end{aligned}\)
putting \(x=0\) and \(y=-2\) we get \(c=2\)
\(\begin{aligned}
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+2 \\
& \Rightarrow y^2-x^2=4
\end{aligned}\)
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