Multiple choice

If the locus of the point of intersection of perpendicular tangents to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is a circle with centre at $\left (0,0\right)$, then the radius of the circle would be

  1. $a+b$
  2. $ab$
  3. $\dfrac b a$
  4. $\sqrt { ({ a }^{ 2 }+{ b }^{ 2 }) } $
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D Correct answer
Explanation

The locus of the intersection of perpendicular tangents to an ellipse is its director circle, which is given by x^2 + y^2 = a^2 + b^2. The radius is sqrt(a^2 + b^2).

AI explanation

The locus of the point of intersection of perpendicular tangents to an ellipse is known as the director circle. For the standard ellipse x^2/a^2 + y^2/b^2 = 1, the equation of its director circle is x^2 + y^2 = a^2 + b^2. Since this locus is given as a circle centered at the origin (0,0), its standard form directly reveals the radius. The radius squared is a^2 + b^2, making the radius sqrt(a^2 + b^2).

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