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Tangents are drawn from the point ($\alpha$,$\beta$) to the hyperbola $3x^2 - 2y^2 = 6 $ and are inclined at angles A and B to the x -axis. If $\tan A. \tan B = 2$, prove that $\beta^2$=2$\alpha^2$ - 7.

I tried the following concept,The points on Hyperbola from where tangent are drawn are $(\sqrt2*(\sec A),\sqrt3*(\tan A))$ & $(\sqrt2*(\sec B),\sqrt3*(\tan B))$

The two tangents are $\frac{\sec A}{\sqrt2}$-$\frac{\tan A}{\sqrt3}$=1 and $\frac{\sec B}{\sqrt2}$-$\frac{\tan B}{\sqrt3}=1$, I tried entering ($\alpha$,$\beta$) and using condition $\tan A. \tan B = 2$ but not able to get the result $\beta^2$=2$\alpha^2$ - 7.

The Dead Legend
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Samar Imam Zaidi
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2 Answers2

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We have $$H:3x^2-2y^2=6$$ and let the tangents be $T:y=mx+c$

$(1):\beta=m\alpha+c,c=\beta-m\alpha$

Sub $T$ into $H$:

$$3x^2-2(mx+c)^2=6$$

$$3x^2-2(m^2x^2+2mcx+c^2)-6=0$$

$$3x^2-2m^2x^2-4mcx-2c^2-6=0$$

$$(3-2m^2)x^2-4mcx-(2c^2+6)=0$$

For tangency, $\Delta=0$

$$16m^2c^2+4(3-2m^2)(2c^2+6)=0$$

$$16m^2c^2+4(6c^2+18-4m^2c^2-12m^2)=0$$

$$16m^2c^2+24c^2+72-16m^2c^2-48m^2=0$$

$$48m^2-24(\beta-m\alpha)^2-72=0$$

$$48m^2-24(\beta^2-2\alpha\beta m+\alpha^2m^2)-72=0$$

$$(48-24\alpha^2)m^2+48\alpha\beta m-(24\beta^2+72)=0$$

$$(2-\alpha^2)m^2+2\alpha\beta m-(\beta^2+3)=0$$

Note that $\tan A$ and $\tan B$ are the slopes of tangents

$(\tan A)(\tan B)=$ product of roots in the equation $=2$

$$-\frac{\beta^2+3}{2-\alpha^2}=2$$

$$\beta^2+3=2\alpha^2-4$$

$$\beta^2=2\alpha^2-7$$

user362325
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  • Thankyou for your prompt answer, i never thought of using quadriatic equation and using sum and product rule of quadriatic equation – Samar Imam Zaidi Aug 26 '17 at 02:18
  • For tangency, Δ=0, I never saw this concept, can you send me link where this concept is mentioned so that I can utilize it for future problem – Samar Imam Zaidi Aug 26 '17 at 02:24
  • @SamarImamZaidi Since we are solving for the intersections of H and T, it should have one root for the quadratiic equation for tangency. – user362325 Aug 26 '17 at 04:00
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The line $y=mx+n$ is a tangent line to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ iff $n^2=m^2a^2-b^2$.

Since our hyperbola it's $$\frac{x^2}{2}-\frac{y^2}{3}=1,$$ we obtaine $n^2=2m^2-3.$

In another hand, $$\beta=\alpha m+n$$ or $$n=\beta-\alpha m.$$ Thus, $$(\beta-\alpha m)^2=2m^2-3$$ or $$(\alpha^2-2)m^2-2\alpha\beta m+\beta^2+3=0$$ and since $m_1m_2=2,$ we obtain $$\frac{\beta^2+3}{\alpha^2-2}=2$$ or $$\beta^2=2\alpha^2-7.$$ Done!

Michael Rozenberg
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