Question:
The common tangent of the parabolas $y^2 = 4x$ and $x^2 = -8y$ is
The common tangent of the parabolas $y^2 = 4x$ and $x^2 = -8y$ is
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Use slope form directly for common tangent problems.
Updated On: Apr 23, 2026
- $y = x + 2$
- $y = x - 2$
- $y = 2x + 3$
- None of these
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The Correct Option is D
Solution and Explanation
Concept:
Tangent in slope form for parabola
Step 1: Tangent to $y^2 = 4x$ is \[ y = mx + \frac{1}{m} \]
Step 2: Tangent to $x^2 = -8y$ is \[ y = mx - 2m^2 \]
Step 3: Common tangent $\Rightarrow$ equate both: \[ mx + \frac{1}{m} = mx - 2m^2 \]
Step 4: \[ \frac{1}{m} = -2m^2 \Rightarrow 1 = -2m^3 \] \[ m^3 = -\frac{1}{2} \]
Step 5: Value of $m$ is not simple integer, so no option matches.
Conclusion:
Answer = None of these
Step 1: Tangent to $y^2 = 4x$ is \[ y = mx + \frac{1}{m} \]
Step 2: Tangent to $x^2 = -8y$ is \[ y = mx - 2m^2 \]
Step 3: Common tangent $\Rightarrow$ equate both: \[ mx + \frac{1}{m} = mx - 2m^2 \]
Step 4: \[ \frac{1}{m} = -2m^2 \Rightarrow 1 = -2m^3 \] \[ m^3 = -\frac{1}{2} \]
Step 5: Value of $m$ is not simple integer, so no option matches.
Conclusion:
Answer = None of these
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