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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

The locus of the point of intersection of two tangents of the...

The locus of the point of intersection of two tangents of the hyperbola x22−y24=1,\dfrac{x^{2}}{2} - \dfrac{y^{2}}{4} = 1, if the product of their slopes is 1,1,& is

Options

  1. Ay2−4=(x2+2)y^{2} - 4 = \left(x^{2} + 2\right)
  2. By2+4=(x2−2)y^{2} + 4 = \left(x^{2} - 2\right)
  3. Cy2+2=(x2−4)y^{2} + 2 = \left(x^{2} - 4\right)
  4. Dy2−2=(x2+4)y^{2} - 2 = \left(x^{2} + 4\right)

Correct answer

B. y2+4=(x2−2)y^{2} + 4 = \left(x^{2} - 2\right)

Step-by-step solution

Equation of any tangent of the hyperbola with slope mm is y=mx±2m2−4y = mx \pm \sqrt{2m^{2} - 4}
If it passes through \left({x}_{1}, {y}_{1}\right)then<br/>(y1−mx1)2=2m2−4⇒(x12−2)m2−2x1y1m+(y12+4)=0then<br />\left({y}_{1} - m{x}_{1}\right)2 = 2m^{2} - 4 ⇒ \left({x}_{1}^{2} - 2\right)m^{2} - 2{x}_{1}{y}_{1}m + \left({y}_{1}^{2} + 4\right) = 0
If n=m1,m2n = {m}_{1}, {m}_{2} then as given m1m2=1{m}_{1}{m}_{2} = 1

⇒y12+4x12−2=1⇒ \dfrac{{y}_{1}^{2} + 4}{{x}_{1}^{2} - 2} = 1
Hence, the required locus is y2+4=(x2−2)y^{2} + 4 = \left(x^{2} - 2\right)

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