A common tangent to the conics x^2 = 6y and 2 | vedclass

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(A) Given conics are $x^2 = 6y$ $(i)$ and $2x^2 - 4y^2 = 9$ $(ii)$.Consider the line $x - y = \frac{3}{2}$ $(iii)$,which implies $x = y + \frac{3}{2}$

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$x - y = \frac{3}{2}$

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(A) Given conics are $x^2 = 6y$ $(i)$ and $2x^2 - 4y^2 = 9$ $(ii)$.Consider the line $x - y = \frac{3}{2}$ $(iii)$,which implies $x = y + \frac{3}{2}$.Substitute $x = y + \frac{3}{2}$ into $(i)$: $(y + \frac{3}{2})^2 = 6y \implies y^2 + 3y + \frac{9}{4} = 6y \implies y^2 - 3y + \frac{9}{4} = 0 \implies (y - \frac{3}{2})^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the parabola.Substitute $x = y + \frac{3}{2}$ into $(ii)$: $2(y + \frac{3}{2})^2 - 4y^2 = 9 \implies 2(y^2 + 3y + \frac{9}{4}) - 4y^2 = 9 \implies 2y^2 + 6y + \frac{9}{2} - 4y^2 = 9 \implies -2y^2 + 6y - \frac{9}{2} = 0 \implies 4y^2 - 12y + 9 = 0 \implies (2y - 3)^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the hyperbola.Therefore,$x - y = \frac{3}{2}$ is the common tangent.

$A$ common tangent to the conics $x^2 = 6y$ and $2x^2 - 4y^2 = 9$ is

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