JEE Main & Advanced Mathematics Conic Sections Rectangular or Equilateral Hyperbola

Rectangular or Equilateral Hyperbola

Category : JEE Main & Advanced

 

(1) Definition : A hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The eccentricity of rectangular hyperbola is always \[\sqrt{2}\].

 

 

The general equation of second degree represents a rectangular hyperbola if \[\Delta \ne 0,\,\,{{h}^{2}}>ab\] and  coefficient of \[{{x}^{2}}+\] coefficient of \[{{y}^{2}}=0\].

 

 

(2) Parametric co-ordinates of a point on the hyperbola \[\mathbf{XY=}{{\mathbf{c}}^{\mathbf{2}}}\] : If \[t\] is non–zero variable, the coordinates of any point on the rectangular hyperbola \[xy={{c}^{2}}\] can be written as \[(ct,c/t)\]. The point \[(ct,c/t)\] on the hyperbola \[xy={{c}^{2}}\] is generally referred as the point \['t'\].

 

 

For rectangular hyperbola the coordinates of foci are \[(\pm a\sqrt{2},\,0)\] and directrices are \[x=\pm a\sqrt{2}\].

 

 

For rectangular hyperbola \[xy={{c}^{2}},\]  the coordinates of foci are \[{{a}^{2}}{{l}^{2}}+{{b}^{2}}{{m}^{2}}={{n}^{2}}\] and directrices are \[x+y=\pm c\sqrt{2}\].

 

 

(3) Equation of the chord joining points \[{{\mathbf{t}}_{\mathbf{1}}}\] and \[{{\mathbf{t}}_{\mathbf{2}}}\] : The equation of the chord joining two points \[\left( c{{t}_{1}},\frac{c}{{{t}_{1}}} \right)\,\text{and}\,\left( c{{t}_{2}},\frac{c}{{{t}_{2}}} \right)\]  on the hyperbola \[xy={{c}^{2}}\] is \[y-\frac{c}{{{t}_{1}}}=\frac{\frac{c}{{{t}_{2}}}-\frac{c}{{{t}_{1}}}}{c{{t}_{2}}-c{{t}_{1}}}(x-c{{t}_{1}})\]

 

 

\[\Rightarrow x+y\,{{t}_{1}}{{t}_{2}}=c\,({{t}_{1}}+{{t}_{2}})\].

 

 

(4) Equation of tangent in different forms

 

 

(i) Point form : The equation of tangent at \[({{x}_{1}},{{y}_{1}})\] to the hyperbola \[xy={{c}^{2}}\] is \[x{{y}_{1}}+y{{x}_{1}}=2{{c}^{2}}\] or \[\frac{x}{{{x}_{1}}}+\frac{y}{{{y}_{_{1}}}}=2\].

 

 

(ii) Parametric form :  The equation of the tangent at \[\left( ct,\frac{c}{t} \right)\] to the hyperbola \[xy={{c}^{2}}\] is \[\frac{x}{t}+yt=2c\].On replacing \[{{x}_{1}}\] by \[ct\] and \[{{y}_{1}}\] by \[\frac{c}{t}\] on the equation of the tangent at \[({{x}_{1}},{{y}_{1}})\]

 

 

i.e., \[x{{y}_{1}}+y{{x}_{1}}=2{{c}^{2}}\] we get \[\frac{x}{t}+yt=2c\].

 

 

Point of intersection of tangents at \['{{t}_{1}}'\] and \['{{t}_{2}}'\] is \[\left( \frac{2c{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}},\,\frac{2c}{{{t}_{1}}+{{t}_{2}}} \right)\].

 

 

(5) Equation of the normal in different forms :

 

 

(i) Point form : The equation of the normal at \[({{x}_{1}},{{y}_{1}})\] to the hyperbola \[xy={{c}^{2}}\] is \[x{{x}_{1}}-y{{y}_{1}}=x_{1}^{2}-y_{1}^{2}\].

 

 

(ii) Parametric form : The equation of the normal at \[\left( ct,\frac{c}{t} \right)\] to the hyperbola \[xy={{c}^{2}}\] is \[x{{t}^{3}}-yt-c{{t}^{4}}+c=0\].

 

 

On replacing \[{{x}_{1}}\] by \[ct\] and \[{{y}_{1}}\] by \[c/t\] in the equation.

 

 

We obtain \[x{{x}_{1}}-y{{y}_{1}}=x_{1}^{2}-y_{1}^{2},\]

 

 

 \[xct-\frac{yc}{t}={{c}^{2}}{{t}^{2}}-\frac{{{c}^{2}}}{{{t}^{2}}}\Rightarrow x{{t}^{3}}-yt-c{{t}^{4}}+c=0\].

 

 

This equation is a fourth degree in \[t\]. So, in general four normals can be drawn from a point to the hyperbola \[xy={{c}^{2}}\], and point of intersection of normals at \[{{t}_{1}}\] and \[{{t}_{2}}\] is

 

 

\[\left( \frac{c\,\{{{t}_{1}}{{t}_{2}}(t_{1}^{2}+{{t}_{1}}{{t}_{2}}+t_{2}^{2})-1\}}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})},\,\,\frac{c\,\{t_{1}^{3}t_{2}^{3}+(t_{1}^{2}+{{t}_{1}}{{t}_{2}}+{{t}_{2}})\}}{{{t}_{1}}{{t}_{2}}({{t}_{1}}+{{t}_{2}})} \right)\].

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