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Find the equation of line touching both the parabolas

$$ x^2=-32y.......(1)$$ $$ y^2=4x.........(2) $$

i have equated slopes of both the parabolas and applied the condition that all the points on the line lie outside the parabolas.

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HINT:

Parametric equation of $x^2=-32y: x=8t,y=-2t^2$

The equation of tangent at $(8t,-2t^2)$ will be $$x(8t)=-16(y-2t^2)\iff xt+2y-4t^2=0$$

Similarly find the equation of tangent of $y^2=4x$ at $(v^2,2v)$

These two equation must be same$\implies$

the ratio of the coefficients of $x$

$=$ the ratio of the coefficients of $y$

$=$ the ratio of the constants

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  • $\begingroup$ can you tell me how did you get the equation of tangent ? Thanks $\endgroup$ – ABC123 Apr 25 '16 at 9:12
  • $\begingroup$ Use slope-point form of an equation to get..$\frac{y-(-2t^2)}{x-8t}$=$\frac{dy}{dx}$. where $x^2 = -32y$ $\endgroup$ – Varun Kumar Apr 25 '16 at 10:03
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General equation for a tangent of the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$.And similarly for the parabola $x^2=4by$ the general equation of tangent is $x=ny+\frac{b}{n}$.According to the question $a=1$ and $b=-8$.And by comparing the equations $y=mx+\frac{1}{m}$ and $x=ny-\frac{8}{n}$(which have to be the same) we can conclude that $m=0.5$ and $n=2$.Therefore the equation of the tangent is :- $$x=2y-4$$I hope this was helpful.

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Let the common tangent line be $y=mx+q$. Then we know that the resolvent equations of the systems $$ \begin{cases} y=mx+q\\ x^2=-32y \end{cases} \qquad \begin{cases} y=mx+q\\ y^2=4x \end{cases} $$ have zero discriminant. The resolvent equations are $$ x^2+32mx+32q=0 \qquad m^2x^2+(2mq-4)x+q^2=0 $$ so we get $$ \begin{cases} 32^2m^2-4\cdot32q=0\\[4px] (2mq-4)^2-4m^2q^2=0 \end{cases} $$ that simplifies to $$ \begin{cases} q=8m^2 \\[4px] mq-1=0 \end{cases} $$ Can you finish?

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