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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
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 x2a2−y2b2=1\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1 which passes through a fixed point (α,β)\left(\alpha, \beta\right) is hyperbola whose centre is -

Options

  1. A(α,β)\left(\alpha, \beta\right)
  2. B(2α,2β)\left(2\alpha, 2\beta\right)
  3. C(α/2,β/2)\left(\alpha / 2, \beta / 2\right)
  4. DNone of these

Correct answer

C. (α/2,β/2)\left(\alpha / 2, \beta / 2\right)

Step-by-step solution

T=S1T = {S}_{1} at mid point (h,k)hxa2−kyb2=h2a2−k2b2\left(h, k\right)\dfrac{hx}{a^{2}} - \dfrac{ky}{b^{2}} = \dfrac{h^{2}}{a^{2}} - \dfrac{k^{2}}{b^{2}}

pass \left(\alpha, \beta\right)point<br/><br/>point<br/><br/>∴ \dfrac{h\alpha}{a^{2}} – \dfrac{ky}{b^{2}} = \dfrac{h^{2}}{a^{2}} – \dfrac{k^{2}}{b^{2}}<br/><br/><br/><br/>∴locusislocus is\dfrac{x^{2} - \alphax}{a^{2}} - \dfrac{y^{2} - \betay}{b^{2}} = 0<br/><br/><br/><br/>⇒ \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)

=k2(let)= k^{2} \left(let\right)

is a hyperbola.

Where centre is (α/2,β/2)\left(\alpha/ 2, \beta/ 2\right) .
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