Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade None
Question:
Which of the following pair $(s)$ is/are orthogonal?
$16x^2+y^2 = C$ and $y^{10} = kx$
$y = x+Ce^{-x}$ and $x+2 = y+ke^{-y}$
$y = Cx^2$ and $x^2+2y^2 = k$
$x^2-y^2 = C$ and $xy = k$
Step-by-Step Solution
Key Concept: Two curves are orthogonal if the product of their slopes at intersection equals $-1$.
Two curves are orthogonal if their tangents are perpendicular at intersection points, meaning $m_1 \cdot m_2 = -1$. For option 1: From $16x^2+y^2=C$, we get $m_1=-\frac{32x}{y}$. From $y^{10}=kx$, we get $m_2=\frac{10y^9}{k}$. At intersection, $m_1 \cdot m_2 = -\frac{32x}{y} \cdot \frac{10y^9}{k} = -1$ when $k=320xy^8$. For option 2: Differentiating gives $m_1=1$ and $m_2=-1$, so $m_1 \cdot m_2=-1$. For option 3: $m_1=2Cx$ and $m_2=-\frac{x}{2y}$, giving $m_1 \cdot m_2=-1$ when $C=\frac{1}{4y^2}$. For option 4: $m_1=\frac{y}{x}$ and $m_2=-\frac{y}{x}$, so $m_1 \cdot m_2=-1$ always.
Correct Answer: 1,2,3,4