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Question: Prove that the angles subtended at its vertices by any chord which is parallel to its conjugate axis...

Prove that the angles subtended at its vertices by any chord which is parallel to its conjugate axis are supplementary.

Answer

The angles subtended at the vertices by any chord which is parallel to its conjugate axis are supplementary for a rectangular hyperbola.

Explanation

Solution

Let the equation of the rectangular hyperbola be x2a2−y2a2=1\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1, or x2−y2=a2x^2 - y^2 = a^2.

The vertices are A1=(a,0)A_1 = (a, 0) and A2=(−a,0)A_2 = (-a, 0).

Let the chord parallel to the conjugate axis (y-axis) be x=cx=c, where ∣c∣>a|c| > a.

The points of intersection of the chord with the hyperbola are P1=(c,y0)P_1 = (c, y_0) and P2=(c,−y0)P_2 = (c, -y_0), where y0=c2−a2y_0 = \sqrt{c^2 - a^2}.

Let θ1\theta_1 be the angle subtended by the chord P1P2P_1P_2 at vertex A1=(a,0)A_1=(a, 0). This is the angle between the vectors A1P1⃗=(c−a,y0)\vec{A_1 P_1} = (c-a, y_0) and A1P2⃗=(c−a,−y0)\vec{A_1 P_2} = (c-a, -y_0).

cos⁡θ1=A1P1⃗⋅A1P2⃗∣A1P1⃗∣∣A1P2⃗∣=(c−a)2−y02(c−a)2+y02(c−a)2+(−y0)2=(c−a)2−y02(c−a)2+y02\cos \theta_1 = \frac{\vec{A_1 P_1} \cdot \vec{A_1 P_2}}{|\vec{A_1 P_1}| |\vec{A_1 P_2}|} = \frac{(c-a)^2 - y_0^2}{\sqrt{(c-a)^2 + y_0^2}\sqrt{(c-a)^2 + (-y_0)^2}} = \frac{(c-a)^2 - y_0^2}{(c-a)^2 + y_0^2}.

Substitute y02=c2−a2y_0^2 = c^2 - a^2:

cos⁡θ1=(c−a)2−(c2−a2)(c−a)2+(c2−a2)=c2−2ac+a2−c2+a2c2−2ac+a2+c2−a2=2a2−2ac2c2−2ac=2a(a−c)2c(c−a)\cos \theta_1 = \frac{(c-a)^2 - (c^2 - a^2)}{(c-a)^2 + (c^2 - a^2)} = \frac{c^2 - 2ac + a^2 - c^2 + a^2}{c^2 - 2ac + a^2 + c^2 - a^2} = \frac{2a^2 - 2ac}{2c^2 - 2ac} = \frac{2a(a-c)}{2c(c-a)}.

Since ∣c∣>a|c|>a, c≠ac \ne a, so c−a≠0c-a \ne 0.

cos⁡θ1=a(a−c)c(c−a)=−ac\cos \theta_1 = \frac{a(a-c)}{c(c-a)} = -\frac{a}{c}.

Let θ2\theta_2 be the angle subtended by the chord P1P2P_1P_2 at vertex A2=(−a,0)A_2=(-a, 0). This is the angle between the vectors A2P1⃗=(c−(−a),y0)=(c+a,y0)\vec{A_2 P_1} = (c-(-a), y_0) = (c+a, y_0) and A2P2⃗=(c−(−a),−y0)=(c+a,−y0)\vec{A_2 P_2} = (c-(-a), -y_0) = (c+a, -y_0).

cos⁡θ2=A2P1⃗⋅A2P2⃗∣A2P1⃗∣∣A2P2⃗∣=(c+a)2−y02(c+a)2+y02(c+a)2+(−y0)2=(c+a)2−y02(c+a)2+y02\cos \theta_2 = \frac{\vec{A_2 P_1} \cdot \vec{A_2 P_2}}{|\vec{A_2 P_1}| |\vec{A_2 P_2}|} = \frac{(c+a)^2 - y_0^2}{\sqrt{(c+a)^2 + y_0^2}\sqrt{(c+a)^2 + (-y_0)^2}} = \frac{(c+a)^2 - y_0^2}{(c+a)^2 + y_0^2}.

Substitute y02=c2−a2y_0^2 = c^2 - a^2:

cos⁡θ2=(c+a)2−(c2−a2)(c+a)2+(c2−a2)=c2+2ac+a2−c2+a2c2+2ac+a2+c2−a2=2a2+2ac2c2+2ac=2a(a+c)2c(c+a)\cos \theta_2 = \frac{(c+a)^2 - (c^2 - a^2)}{(c+a)^2 + (c^2 - a^2)} = \frac{c^2 + 2ac + a^2 - c^2 + a^2}{c^2 + 2ac + a^2 + c^2 - a^2} = \frac{2a^2 + 2ac}{2c^2 + 2ac} = \frac{2a(a+c)}{2c(c+a)}.

Since ∣c∣>a|c|>a, c≠−ac \ne -a, so c+a≠0c+a \ne 0.

cos⁡θ2=a(a+c)c(c+a)=ac\cos \theta_2 = \frac{a(a+c)}{c(c+a)} = \frac{a}{c}.

We have cos⁡θ1=−a/c\cos \theta_1 = -a/c and cos⁡θ2=a/c\cos \theta_2 = a/c.

Thus, cos⁡θ1=−cos⁡θ2\cos \theta_1 = -\cos \theta_2.

Since θ1\theta_1 and θ2\theta_2 are angles subtended by a chord, they are angles within triangles A1P1P2A_1 P_1 P_2 and A2P1P2A_2 P_1 P_2. Thus 0<θ1<π0 < \theta_1 < \pi and 0<θ2<π0 < \theta_2 < \pi.

The condition cos⁡θ1=−cos⁡θ2\cos \theta_1 = -\cos \theta_2 for angles in (0,π)(0, \pi) implies θ1=π−θ2\theta_1 = \pi - \theta_2, which means θ1+θ2=π\theta_1 + \theta_2 = \pi.

Therefore, the angles subtended at the vertices by any chord which is parallel to the conjugate axis are supplementary, provided the hyperbola is rectangular.

If the question is intended for a general hyperbola, the statement is false. Assuming the question is valid and expects a proof, it must be implicitly referring to a rectangular hyperbola.