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If the tangent drawn to the hyperbola \mathrm{4 y^2=x^2+1} intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is

Option: 1

\mathrm{4 x^2-y^2-16 x^2 y^2=0}


Option: 2

\mathrm{4 x^2-y^2+16 x^2 y^2=0}


Option: 3

\mathrm{x^2-4 y^2+16 x^2 y^2=0}


Option: 4

\mathrm{x^2-4 y^2-16 x^2 y^2=0}


Answers (1)

best_answer

We have, \mathrm{4 y^2=x^2+1}                               ...(i)

The tangent to (i) at \mathrm{\left(x_1, y_1\right)} is given by \mathrm{4 y y_1=x x_1+1}

According to question, \mathrm{A \equiv\left(\frac{-1}{x_1}, 0\right), B \equiv\left(0, \frac{1}{4 y_1}\right)}

Let mid point of AB be M(h, k).

Then, \mathrm{\frac{-1}{x_1}=2 h \Rightarrow x_1=\frac{-1}{2 h}}

and \mathrm{\frac{1}{4 y_1}=2 k \Rightarrow y_1=\frac{1}{8 k}}

\mathrm{\left(x_1, y_1\right)} lies on (i)

\mathrm{ \therefore \quad 4\left(\frac{1}{8 k}\right)^2=\left(\frac{-1}{2 h}\right)^2+1 \Rightarrow \frac{4 \times 1}{64 k^2}=\frac{1}{4 h^2}+1 }

\mathrm{ \Rightarrow \frac{1}{16 k^2}=\frac{1}{4 h^2}+1 \Rightarrow h^2=4 k^2+16 h^2 k^2 }

\therefore \quad  Locus of mid point of AB is

\mathrm{ x^2=4 y^2+16 x^2 y^2 \text { or } x^2-4 y^2-16 x^2 y^2=0 }

Posted by

Kuldeep Maurya

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