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  • The locus of the point of intersection of two perpendicular tangents to the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cm

The locus of the point of intersection of two perpendicular tangents to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} is

Option: 1

x^2+y^2=a^2-b^2


Option: 2

x^2+y^2=a^2


Option: 3

x^2+y^2=b^2


Option: 4

x^2+y^2=a^2+b^2


Answers (1)

best_answer

Equation of any tangent in terms of slope m is

y=m x+\sqrt{\left(a^2 m^2-b^2\right)}

It passes through (h, k), then (k-m h)^2=a^2 m^2-b^2

\Rightarrow m^2\left(h^2-a^2\right)-2 m h k+k^2+b^2=0
which is quadratic in m.

Let slopes of tangents are m_1 and m_2, then m_1 m_2=-1

\Rightarrow \frac{k^2+b^2}{h^2-a^2}=-1 \Rightarrow h^2+k^2=a^2-b^2

Hence, required locus is x^2+y^2=a^2-b^2 which is director circle of \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

Posted by

Pankaj Sanodiya

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