Question:
Equation of the line passing through the points of intersection of the parabola \(x^2=8y\) and the ellipse \(\frac{x^2}{3}+y^2=1\) is :
Equation of the line passing through the points of intersection of the parabola \(x^2=8y\) and the ellipse \(\frac{x^2}{3}+y^2=1\) is :
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Substitute x^2 = 8y into the ellipse to find the common y-coordinates, rejecting any that give no real x.
Updated On: Oct 31, 2026
- \(y-3=0\)
- \(y+3=0\)
- \(3y+1=0\)
- \(3y-1=0\)
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The Correct Option is D
Solution and Explanation
Put \(x^2=8y\) in \(\frac{x^2}{3}+y^2=1\): \(\frac{8y}{3}+y^2=1\Rightarrow 3y^2+8y-3=0\).
\((3y-1)(y+3)=0\Rightarrow y=\frac13\) or \(y=-3\).
For \(y=-3\), \(x^2=-24\lt0\), which is not real, so rejected.
For \(y=\frac13\), \(x^2=\frac83\), giving two real points \(\left(\pm\sqrt{8/3},\ \frac13\right)\).
The line through both points is \(y=\frac13\), i.e. \(3y-1=0\).
Answer: (4) \(3y-1=0\)
\((3y-1)(y+3)=0\Rightarrow y=\frac13\) or \(y=-3\).
For \(y=-3\), \(x^2=-24\lt0\), which is not real, so rejected.
For \(y=\frac13\), \(x^2=\frac83\), giving two real points \(\left(\pm\sqrt{8/3},\ \frac13\right)\).
The line through both points is \(y=\frac13\), i.e. \(3y-1=0\).
Answer: (4) \(3y-1=0\)
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