Normal are drawn to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`
at point `theta_1` and `theta_2`
meeting the conjugate axis at `G_1a n dG_2,`
respectively. If `theta_1+theta_2=pi/2,`
prove that `C G_1*C G_2=(a^2e^4)/(e^2-1)`
, where `C`
is the center of the hyperbola and `e`
is the eccentricity.
Text Solution
AI Generated Solution
To solve the problem, we need to prove that \( C G_1 \cdot C G_2 = \frac{a^2 e^4}{e^2 - 1} \), where \( C \) is the center of the hyperbola, and \( e \) is the eccentricity.
### Step-by-Step Solution:
1. **Identify the Hyperbola and Parameters**:
The hyperbola is given by the equation:
\[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
...
Normal are drawn to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 at point theta_1a n dtheta_2 meeting the conjugate axis at G_1a n dG_2, respectively. If theta_1+theta_2=pi/2, prove that C G_1dotC G_2=(a^2e^4)/(e^2-1) , where C is the center of the hyperbola and e is the eccentricity.
If e is the eccentricity of the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and theta is the angle between the asymptotes, then cos.(theta)/(2) is equal to
If the normal at P(theta) on the hyperbola (x^2)/(a^2)-(y^2)/(2a^2)=1 meets the transvers axis at G , then prove that A GdotA^(prime)G=a^2(e^4sec^2theta-1) , where Aa n dA ' are the vertices of the hyperbola.
Tangents to the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 make angle theta_(1), theta_(2) with transvrse axis of a hyperbola. Show that the points of intersection of these tangents lies on the curve 2xy=k(x^(2)-a^(2)) when tan theta_(1)+ tan theta_(2)=k
Show that the acute angle between the asymptotes of the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1,(a^2> b^2), is 2cos^(-1)(1/e), where e is the eccentricity of the hyperbola.
If any line perpendicular to the transverse axis cuts the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 and the conjugate hyperbola (x^2)/(a^2)-(y^2)/(b^2)=-1 at points Pa n dQ , respectively, then prove that normal at Pa n dQ meet on the x-axis.
If any line perpendicular to the transverse axis cuts the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 and the conjugate hyperbola (x^2)/(a^2)-(y^2)/(b^2)=-1 at points Pa n dQ , respectively, then prove that normal at Pa n dQ meet on the x-axis.
The hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 passes through the point (2, ) and has the eccentricity 2. Then the transverse axis of the hyperbola has the length 1 (b) 3 (c) 2 (d) 4
If the normal at 'theta' on the hyperbola (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 meets the transverse axis at G , and A and A' are the vertices of the hyperbola , then AC.A'G= (a) a^2(e^4 sec^2 theta-1) (b) a^2(e^4 tan^2 theta-1) (c) b^2(e^4 sec^2 theta-1) (d) b^2(e^4 sec^2 theta+1)
If the normals at P(theta) and Q(pi/2+theta) to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 meet the major axis at Ga n dg, respectively, then P G^2+Qg^2= b^2(1-e^2)(2-e)^2 a^2(e^4-e^2+2) a^2(1+e^2)(2+e^2) b^2(1+e^2)(2+e^2)
