JEE Main
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- JEE Main
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- Mathematics
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- Hyperbola
JEE Main · Mathematics · Hyperbola
Locus of the middle points of the chords of the hyperbola x^ 2 a^ 2...
Locus of the middle points of the chords of the hyperbola which passes through a fixed point is hyperbola whose centre is -
Options
- A
- B
- C
- DNone of these
Correct answer
C.
Step-by-step solution
at mid point
pass \left(\alpha, \beta\right)∴ \dfrac{h\alpha}{a^{2}} – \dfrac{ky}{b^{2}} = \dfrac{h^{2}}{a^{2}} – \dfrac{k^{2}}{b^{2}}∴\dfrac{x^{2} - \alphax}{a^{2}} - \dfrac{y^{2} - \betay}{b^{2}} = 0⇒ \dfrac{{\left(x - \alpha / 2\right)}^{2}}{a^{2}} - \dfrac{{\left(y - \beta / 2\right)}^{2}}{b^{2}} = \dfrac{1}{4} \left(\dfrac{{\alpha}^{2}}{a^{2}} - \dfrac{{\beta}^{2}}{b^{2}}\right)
is a hyperbola.
Where centre is .
pass \left(\alpha, \beta\right)∴ \dfrac{h\alpha}{a^{2}} – \dfrac{ky}{b^{2}} = \dfrac{h^{2}}{a^{2}} – \dfrac{k^{2}}{b^{2}}∴\dfrac{x^{2} - \alphax}{a^{2}} - \dfrac{y^{2} - \betay}{b^{2}} = 0⇒ \dfrac{{\left(x - \alpha / 2\right)}^{2}}{a^{2}} - \dfrac{{\left(y - \beta / 2\right)}^{2}}{b^{2}} = \dfrac{1}{4} \left(\dfrac{{\alpha}^{2}}{a^{2}} - \dfrac{{\beta}^{2}}{b^{2}}\right)
is a hyperbola.
Where centre is .