Two parabolas with a common vertex at the ori | vedclass

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(C) Since the origin is the common vertex,let the equations of the two parabolas be $y^2 = 4ax$ and $x^2 = 4by$.Given the length of the latus rectum i

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$4(x+y)+3 = 0$

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(C) Since the origin is the common vertex,let the equations of the two parabolas be $y^2 = 4ax$ and $x^2 = 4by$.Given the length of the latus rectum is $3$,we have $4a = 3$ and $4b = 3$,so $a = b = \frac{3}{4}$.The equations are $y^2 = 3x$ and $x^2 = 3y$.Let the common tangent be $y = mx + c$.For $y^2 = 3x$,substituting $y = mx + c$ gives $(mx + c)^2 = 3x$,or $m^2x^2 + (2mc - 3)x + c^2 = 0$.Since it is a tangent,the discriminant is zero: $(2mc - 3)^2 - 4m^2c^2 = 0$,which simplifies to $9 - 12mc = 0$,so $c = \frac{3}{4m}$.For $x^2 = 3y$,substituting $y = mx + c$ gives $x^2 = 3(mx + c)$,or $x^2 - 3mx - 3c = 0$.Setting the discriminant to zero: $(-3m)^2 - 4(1)(-3c) = 0$,which gives $9m^2 + 12c = 0$,so $c = -\frac{3m^2}{4}$.Equating the two expressions for $c$: $\frac{3}{4m} = -\frac{3m^2}{4}$ $\Rightarrow m^3 = -1$ $\Rightarrow m = -1$.Then $c = \frac{3}{4(-1)} = -\frac{3}{4}$.The equation of the tangent is $y = -x - \frac{3}{4}$,which simplifies to $4(x + y) + 3 = 0$.

Two parabolas with a common vertex at the origin and with axes along the $x-$ axis and $y-$ axis,respectively,intersect each other in the first quadrant. If the length of the latus rectum of each parabola is $3$,then the equation of the common tangent to the two parabolas is?

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