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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

Let \(a\) and \(b\) be any two numbers satisfying \(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{4}\). Then, the foot of perpendicular from the origin on the variable line, \(\frac{x}{a} + \frac{y}{b} = 1\) , lies on

  1. A a hyperbola with each semi-axis \( = \sqrt 2 \)
  2. B a hyperbola with each semi-axis \(= 2\)
  3. C a circle of radius \(= 2\)
  4. D a circle of radius \( = \sqrt 2 \)
Verified Solution

Answer & Solution

Correct Answer

(C) a circle of radius \(= 2\)

Step-by-step Solution

Detailed explanation

Let the foot of the perpendicular from \((0,0)\) on the variable line \(\frac{x}{a} + \frac{y}{b} = 1\,\) is \(\left( {{x_1} > {y_1}} \right)\) Hence, perpendicular distance of the variable line \(\frac{x}{a} + \frac{y}{b} = 1\,\,\) from the point \(O\)…
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