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  • The locus of the middle points of the chords of 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%7Bx%5E2%7D%7

The locus of the middle points of the chords of hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}, which subtend a right angle at the origin, is
\mathrm{ \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^2\left(\frac{1}{a^2}+\frac{\lambda}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4}, \text { where } \lambda= }

Option: 1

0


Option: 2

-1


Option: 3

1


Option: 4

2


Answers (1)

Let \mathrm{P(h, k)} be the middle point of the chord of hyperbola

\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 }                    ......(1)

The equation of the chord is \mathrm{T=S_1.}

\mathrm{ \Rightarrow \quad \frac{h x}{a^2}-\frac{k y}{b^2}-1=\frac{h^2}{a^2}-\frac{k^2}{b^2}-1 }     ......(2)

Using homogeneity method, the combined equation of \mathrm{O A, O B} is

\mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=\left(\frac{\frac{h x}{a^2}-\frac{k y}{b^2}}{\frac{h^2}{a^2}-\frac{k^2}{b^2}}\right)^2 }                      ......(3)

As \mathrm{O A \perp O B}, the condition is that the sum of coefficients of \mathrm{ x^2} and \mathrm{ y^2} in equation (3) is zero

\mathrm{ \begin{aligned} & \Rightarrow \quad \frac{1}{a^2}\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}\right)^2-\frac{1}{b^2}\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}\right)^2-\frac{h^2}{a^4}-\frac{k^2}{b^4}=0 \\\\ & \text { Or } \quad\left(\frac{h^2}{a^2}-\frac{k^2}{b^2}\right)^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=\frac{h^2}{a^4}+\frac{k^2}{b^4} \end{aligned} }
\mathrm{ \therefore \quad}  The locus of \mathrm{ (h, k)} is

\mathrm{ \begin{aligned} & \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)^2\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4} \\ \Rightarrow & \lambda=-1 \end{aligned} }

The answer is (b) 

Posted by

Sumit Saini

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