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  • The locus of a point, from where tangents to the rectangular hyperbola <img alt="\mathrm{ x^2-y^2=a^2 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20x%5E2-y%5E2%3Da%5E2

The locus of a point, from where tangents to the rectangular hyperbola \mathrm{ x^2-y^2=a^2 } contain an angle of \mathrm{ 45^{\circ} } is 

Option: 1

\mathrm{ \left(x^2+y^2\right)+a^2\left(x^2-y^2\right)=4 a^4 }


Option: 2

\mathrm{ 2\left(x^2+y^2\right)+4 a^2\left(x^2-y^2\right)=4 a^2 }


Option: 3

\mathrm{ \left(x^2+y^2\right)+4 a^2\left(x^2-y^2\right)=4 a^4 }


Option: 4

\mathrm{ \left(x^2+y^2\right)+a^2\left(x^2-y^2\right)=a^4 }


Answers (1)

Let \mathrm{ y=m x \pm \sqrt{m^2 a^2-a^2} } be two tangent and passes through (h, k) then \mathrm{ (\mathrm{k}-\mathrm{mk})^2=\mathrm{m}^2 \mathrm{a}^2-\mathrm{a}^2 } 
\mathrm{ \Rightarrow \quad \mathrm{m}^2\left(\mathrm{~h}^2-\mathrm{a}^2\right)-2 \mathrm{khm}+\mathrm{k}^2+\mathrm{a}^2=0 }
\mathrm{ \Rightarrow \quad \mathrm{m}_1+\mathrm{m}_2=\frac{2 \mathrm{kh}}{\mathrm{h}^2-\mathrm{a}^2} } and \mathrm{ \mathrm{m}_1 \mathrm{~m}_2=\frac{k^2+a^2}{h^2-a^2} } using \mathrm{ \tan 45^{\circ}=\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2} }
Hence (c) is the correct answer.

Posted by

Sumit Saini

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