Fun with conics

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Let's 'stretch' or scale up the the problem in the Y direction by a factor of 3/2, thus simplifying the given ellipse and parabola into a circle $x^2 + y^2 = 9$ and a parabola $y^2 = 9x = 4 \left(\frac{9}{4}\right)x$ with focus $F = \left(\frac{9}{4},0\right)$

Let $T_{1}(h,k), T_{2}$ be points of tangency. Then foot of perpendicular P from F on the tangent must lie on the Y axis (= tangent at the vertext = the pedal* curve here!)

Another property of Parabola gives X-intercept of tangent = -h and therefore Y-intercept as k/2

From the two similar triangles formed: $OT_{2}P$ and $POF$ :

$\frac{OP}{PF}=\frac{OT_{2}}{OP}$ $\frac{k/2}{\sqrt{(9/4)^2+(k/2)^2}}=\frac{3}{k/2}$ Giving the equation $k^4-36k-729=0$ solving which we get the Y-Intercept k/2 as $\frac{3}{2}\sqrt{2+\sqrt{13}}$

But this is for the 'stretched' problem of circle and another parabola, so let us shrink it back by a factor of 2/3 to get the desired Y-intercept as $\boxed{\sqrt{2+\sqrt{13}}}$

• http://en.wikipedia.org/wiki/Pedal_curve

This one was quite easier.

General equation of tangent of slope $m$ for

• parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$

• ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $y=mx\pm \sqrt{a^2 m^2+b^2}$

So tangents in these cases would be $y=mx+\frac{1}{m}$ and $y=mx\pm \sqrt{9m^2+4}$ .

Comparing both the equations,

$m^2(9m^2+4)=1$

On solving we get $\frac{1}{m}=\sqrt{2+\sqrt{13}}$

I did this by writing equation of tangents of given parabola and ellipse in parametric form and comparing to eliminate assumed parameters.