Equation the condition of tangency of both the conics $$4m\pm \sqrt {9m^2-4}=4\sqrt {1+m^2}$$my problem is easy. Are there any tips to reduce such heavy calculations?
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Have you been introduced to the $3 \times 3$ matrix associated with a conic curve ? – Jean Marie Mar 08 '20 at 18:05
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That's an equation for $m$, which I assume that's the slope of the tangent. Do you have the intercept as well? – Andrei Mar 08 '20 at 18:07
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$9X^2-4Y^2-1=0$ and $Y^2-\frac{X}{2}-\frac1{16}=0$ are the dual conics, and they intersect in $(\frac12,\pm\frac{\sqrt{5}}{4})$, so the two common tangents are $\frac12 x\pm\frac{\sqrt{5}}{4} y+1=0.$ – Jan-Magnus Økland Mar 08 '20 at 18:34
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@JeanMarie not yet – Aditya Mar 09 '20 at 02:17
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@Andrei no, that’s all I have. $m$ is indeed the slope of the tangents – Aditya Mar 09 '20 at 02:18
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The equation of any tangent of the hyperbola at $(3\sec t,2\tan t)$ will be $$x(3\sec t)/9-y(2\tan t)/4=1\iff 2x\tan t -3y\tan t-6=0$$
Now if this has to be a tangent of the given circle, the distance from the center will be same
$$4=\dfrac{|2\sec t(4)+0-6|}{\sqrt{(2\sec t)^2+(3\tan t)^2}}$$
Take square in both sides and use $$\tan ^2t=\sec^2t-1$$ to form a quadratic equation in $\sec t$
lab bhattacharjee
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