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  • If any line perpendicular to the transverse axis cuts the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7B

If any line perpendicular to the transverse axis cuts the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} and the conjugate hyperbola  \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1} at points P and Q, respectively, then normals at P and Q meet on the
 

Option: 1

 x-axis
 


Option: 2

 y-axis
 


Option: 3

 origin
 


Option: 4

 None of these


Answers (1)

best_answer

Let the perpendicular line cuts the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~ at ~P\left(x_1, y_1\right)} and

the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1}  at points \mathrm{Q\left(x_2, y_2\right)}

Normal to the hyperbola \mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1} at point P is

\mathrm{ \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \quad \quad \quad \dots(i) }

Normal to the hyperbola  \mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 } at Q is

\mathrm{ \frac{a^2 x}{x_2}+\frac{b^2 y}{y_2}=a^2+b^2 \quad \quad \quad \dots(ii)}

In (i) and (ii), putting y=0, we get

\mathrm{ x=\frac{a^2+b^2}{a^2} x_1=\frac{a^2+b^2}{a^2} x_2 \quad\left(\therefore x_1=x_2\right)}

Both meet on x-axis.
 

Posted by

SANGALDEEP SINGH

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