If the line 2x + 5y = 12 intersects the ellip | vedclass

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(D) The given line is $2x + 5y = 12$,which implies $x = \frac{12 - 5y}{2}$.Substituting this into the ellipse equation $4x^2 + 5y^2 = 20$:$4\left(\fra

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None of these

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(D) The given line is $2x + 5y = 12$,which implies $x = \frac{12 - 5y}{2}$.Substituting this into the ellipse equation $4x^2 + 5y^2 = 20$:$4\left(\frac{12 - 5y}{2}\right)^2 + 5y^2 = 20$$4\left(\frac{144 - 120y + 25y^2}{4}\right) + 5y^2 = 20$$144 - 120y + 25y^2 + 5y^2 = 20$$30y^2 - 120y + 124 = 0$Dividing by $2$,we get $15y^2 - 60y + 62 = 0$.The discriminant $D = b^2 - 4ac = (-60)^2 - 4(15)(62) = 3600 - 3720 = -120$.Since $D Therefore,the premise of the question is incorrect as there are no intersection points $A$ and $B$.

If the line $2x + 5y = 12$ intersects the ellipse $4x^2 + 5y^2 = 20$ in two distinct points $A$ and $B$,then the mid-point of $AB$ is

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