Question:
Let \( S \) be the family of orthogonal trajectories of the family of curves
\[
2x^2 + y^2 = k, \text{ for } k \in \mathbb{R} \text{ and } k>0.
\]
If \( \ell \in S \) and \( C \) passes through the point \( (1, 2) \), then \( C \) also passes through
Let \( S \) be the family of orthogonal trajectories of the family of curves
\[
2x^2 + y^2 = k, \text{ for } k \in \mathbb{R} \text{ and } k>0.
\]
If \( \ell \in S \) and \( C \) passes through the point \( (1, 2) \), then \( C \) also passes through
Show Hint
To find the orthogonal trajectories, differentiate the given equation implicitly, then solve for the new family by taking the negative reciprocal of the derivative.
Updated On: Dec 11, 2025
- \( (4, -\sqrt{2}) \)
- \( (2, -4) \)
- \( (2, 2\sqrt{2}) \)
- \( (4, 2\sqrt{2}) \)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Equation of the family of curves.
We are given the family of curves \( 2x^2 + y^2 = k \), where \( k \) is a constant. To find the orthogonal trajectories, we differentiate implicitly with respect to \( x \) to get the slope of the tangent line. This will give the equation of the orthogonal trajectories. \[ 4x + 2yy' = 0 \implies y' = -\frac{2x}{y}. \] The family of orthogonal trajectories will have slopes that are the negative reciprocal of this slope.
Step 2: Use the point \( (1, 2) \).
Substitute the point \( (1, 2) \) into the equation of the curve \( 2x^2 + y^2 = k \) to find the value of \( k \): \[ 2(1)^2 + (2)^2 = k \implies k = 8. \] So, the equation of the curve is \( 2x^2 + y^2 = 8 \). Now, we find the equation of the orthogonal trajectory. The orthogonal trajectory will pass through \( (1, 2) \) and the new point will be \( (2, 2\sqrt{2}) \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(C)} \).
We are given the family of curves \( 2x^2 + y^2 = k \), where \( k \) is a constant. To find the orthogonal trajectories, we differentiate implicitly with respect to \( x \) to get the slope of the tangent line. This will give the equation of the orthogonal trajectories. \[ 4x + 2yy' = 0 \implies y' = -\frac{2x}{y}. \] The family of orthogonal trajectories will have slopes that are the negative reciprocal of this slope.
Step 2: Use the point \( (1, 2) \).
Substitute the point \( (1, 2) \) into the equation of the curve \( 2x^2 + y^2 = k \) to find the value of \( k \): \[ 2(1)^2 + (2)^2 = k \implies k = 8. \] So, the equation of the curve is \( 2x^2 + y^2 = 8 \). Now, we find the equation of the orthogonal trajectory. The orthogonal trajectory will pass through \( (1, 2) \) and the new point will be \( (2, 2\sqrt{2}) \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(C)} \).
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