A common tangent to 9x^2 + 16y^2 = 144,y^2 - | vedclass

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(C) The given equations are:$1$) Ellipse: $9x^2 + 16y^2 = 144 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$2$) Parabola: $y^2 = x - 4$$3$) Circle:

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$x = 4$

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(C) The given equations are:$1$) Ellipse: $9x^2 + 16y^2 = 144 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1$$2$) Parabola: $y^2 = x - 4$$3$) Circle: $x^2 + y^2 - 12x + 32 = 0 \Rightarrow (x-6)^2 + y^2 = 4$For the parabola $y^2 = 4a(x-h)$,the tangent is $y = mx + \frac{a}{m} + h$. Here $4a = 1$,so $a = 1/4$ and $h = 4$. Thus,$y = mx + \frac{1}{4m} + 4$.For the circle $(x-6)^2 + y^2 = 2^2$,the distance from center $(6, 0)$ to the line $mx - y + (\frac{1}{4m} + 4) = 0$ is equal to radius $2$:$\frac{|6m - 0 + \frac{1}{4m} + 4|}{\sqrt{m^2 + 1}} = 2$$|6m + \frac{1}{4m} + 4| = 2\sqrt{m^2 + 1}$.Testing $x = 4$ (vertical line,$m 	o \infty$): For the ellipse,$x=4$ is a tangent at $(4, 0)$.For the parabola $y^2 = x-4$,$x=4$ is a tangent at $(4, 0)$.For the circle $(x-6)^2 + y^2 = 4$,at $x=4$,$(4-6)^2 + y^2 = 4$ $\Rightarrow 4 + y^2 = 4$ $\Rightarrow y = 0$. Thus $x=4$ is a tangent at $(4, 0)$.Therefore,$x=4$ is the common tangent.

$A$ common tangent to $9x^2 + 16y^2 = 144$,$y^2 - x + 4 = 0$,and $x^2 + y^2 - 12x + 32 = 0$ is:

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