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The number of normals that can be drawn from a point \mathrm{A\left(x_1, y_1\right)} to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} is

Option: 1

1
 


Option: 2

2
 


Option: 3

4
 


Option: 4

none of these


Answers (1)

best_answer

The equation of the normal at a point \mathrm{( a \sec \theta, b \tan \theta )} on the hyperbola is

         \mathrm{a x \cos \theta+b y \cot \theta=a^2+b^2}

As it is drawn from \mathrm{A\left(x_1, y_1\right)}, the condition is

        \mathrm{ a x_1 \cos \theta+b y_1 \cot \theta=a^2+b^2 }                             ..........(1)

Or \mathrm{a x_1 \frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}+b y_1 \frac{1-\tan ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}}=a^2+b^2}

On putting \mathrm{\tan \frac{\theta}{2}=t}, this equation becomes

\mathrm{ a x_1 \frac{1-t^2}{1+t^2}+b y_1 \frac{1-t^2}{2 t}=a^2+b^2 }

Or \mathrm{\quad b y_1 t^4+2\left(a^2+b^2+a x_1\right)+2\left(a^2+b^2-a x_1\right)-b y_1=0}

Let its roots be \mathrm{t_1, t_2, t_3, t_4.}

These values given by

\mathrm{ t_1=\tan \frac{\theta_1}{2}, t_2=\tan \frac{\theta_2}{2}, t_3=\tan \frac{\theta_3}{2}, t_4=\tan \frac{\theta_4}{2} }

correspond to four feet of normals \mathrm{\left(a \sec \theta_r, b \tan \theta_r\right)}

\mathrm{(r=1,2,3,4), } drawn from \mathrm{A\left(x_1, y_1\right).}

Hence, four normals can be drawn to the hyperbola.

The answer is (c)

Posted by

Kuldeep Maurya

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