0
$\begingroup$

From any point of the hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$, tangents are drawn to another hyperbola which has the same asymptotes. Show that the chord of contact cuts off a constant area from the asymptotes.

Concept

Lets presume that hyperbola are conjugate

Let the two conjugate hyperbolas be represented as

$\ H_1$: $\frac{y^2}{b^2} -\frac{x^2}{a^2}=1$

$\ H_2$: $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$

There Asymptotes are

$\ A_1$: $\frac{y}{b} -\frac{x}{a}=1$

$\ A_2$: $\frac{x}{a} -\frac{y}{b}=1$

Points on $\ H_1$ is $(a\tan(\alpha), b\sec(\alpha))$

Points on $\ H_2$ is $(a\sec(\alpha), b\tan(\alpha))$

Equation of Line tangent to $\ H_1$

$\frac{y\sec(\alpha)}{b}-\frac{x\tan(\alpha)}{a}$=1, it passes through $(a\sec(\alpha), b\tan(\alpha))$

From here i need help. Do we need to find the intersection of the line and asymptote and then find area under curve that needs to be "Constant"

If i enter coordinates of $\ H_2$ in equation of line equation vanishes

$\endgroup$
  • $\begingroup$ There’s nothing in the problem statement at the top about the hyperbolas being conjugate, only that they have the same asymptotes. $\endgroup$ – amd Sep 5 '17 at 3:16
  • $\begingroup$ How do we prove that chord cut off constant area , $\endgroup$ – Samar Imam Zaidi Sep 5 '17 at 3:19
  • $\begingroup$ @amd Conjugate Hyperloba have same asymptote $\endgroup$ – Samar Imam Zaidi Sep 5 '17 at 3:26
  • 1
    $\begingroup$ Yes, but hyperbolas with the same asymptotes are not necessarily conjugate. Be sure that the problem that you’re trying to solve is the one that was posed. As your question stands right now, it’s not. $\endgroup$ – amd Sep 5 '17 at 3:27
  • $\begingroup$ @amd with no assumption i am presuming that hyperbola are conjugate so that solving process becomes easy, i just need to arrive at some conclusion $\endgroup$ – Samar Imam Zaidi Sep 5 '17 at 4:26
2
$\begingroup$

You need to find the line through the points of tangency, its intersections with the common asymptotes and then the area of the resulting triangle. This is a fairly straightforward calculation if you use homogeneous coordinates.

The family of hyperbolas with the same asymptotes as ${x^2\over a^2}-{y^2\over b^2}=1$ is given by the equations ${x^2\over a^2}-{y^2\over b^2}=k$. (For $k=0$, this is the equation of the asymptotes themselves.) In matrix form, these hyperbolas are $H=\operatorname{diag}(1/a^2,-1/b^2,-k)$. The line of contact of the tangents to a hyperbola from a point $(x,y)$ is the polar line of this point, namely, $$\mathbf l=H[x:y:1]=\left[{x\over a^2}:-{y\over b^2}:-k\right].$$ The two asymptotes are $\mathbf m_1=[b:a:0]$ and $\mathbf m_2=[b:-a:0]$, and their intersections with $\mathbf l$ are $$ \mathbf p_1=\mathbf l\times\mathbf m_1=\left[ak:-bk:\frac xa+\frac yb \right] \\ \mathbf p_2=\mathbf l\times\mathbf m_2=\left[-ak:-bk:-\frac xa+\frac yb\right] $$ which in Cartesian coordinates are, respectively, $$ P_1=\left({a^2bk\over bx+ay},-{ab^2k\over bx+ay}\right) \\ P_2=\left({a^2bk\over bx-ay},{ab^2k\over bx-ay}\right).$$ The area of $\triangle{OP_1P_2}$ is one half of the absolute value of $${a^2bk\over bx+ay}{ab^2k\over bx-ay}+{ab^2k\over bx+ay}{a^2bk\over bx-ay} = 2{a^3b^3k^2\over b^2x^2-a^2y^2}.$$ Multiplying the equation of the original hyperbola by $a^2b^2$ gives the equivalent equation $b^2x^2-a^2y^2=a^2b^2$, so for any point on this hyperbola the area of $\triangle{OP_1P_2}$ is equal to $abk^2$.

Observe that this didn’t use the points of tangency directly but instead used the polar line of a point, so this identity also holds when the point is interior to the other hyperbola, in which case there are no (real) tangent lines, and when the two hyperbolas coincide, in which case there’s only one point of tangency. Also, for any hyperbola in the family $b^2x^2-a^2y^2$ is constant, therefore this constant-area property holds for any pair of hyperbolas with common asymptotes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thanks for your explanation $\endgroup$ – Samar Imam Zaidi Sep 5 '17 at 8:12
  • $\begingroup$ @SamarImamZaidi Another approach would be to prove it for a simple case, such as the hyperbola $xy=1$. The equation for the line of contact is fairly easy to derive and as the asymptotes are the coordinate axes, the intersections are trivial to find. Then, argue that any hyperbola of the form in the problem can be obtained from this one via a linear transformation, which multiples areas by a constant factor. $\endgroup$ – amd Sep 5 '17 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.