Question

From a point P two tangents are drawn to the ellipse x²a²+y²b²=1. If these tangents intersect the coordinate axes at concyclic points, then the locus of the point P is (a>b)

A. x²+y²=a²+b²
B. x²+y²=a²−b²
C. x²−y²=a²−b²
D. x²=y²

Answer 1

The correct answer is option C. The locus of the point P is x²−y²=a²−b². When two tangents are drawn from a point P to the ellipse and they intersect the coordinate axes at concyclic points, the sum of the distances from P to the two foci of the ellipse should be constant.

Step-by-step explanation:

The locus of the point P is C. x²−y²=a²−b².

To understand why, let's consider the given ellipse equation: x²/a² + y²/b² = 1.

When two tangents are drawn from a point P to the ellipse and they intersect the coordinate axes at concyclic points, the sum of the distances from P to the two foci of the ellipse should be constant. In this case, the foci are located at (±c, 0), where c is a positive constant related to a and b by the equation c² = a² - b².

If we let P(x, y) be the point on the ellipse and F₁(c, 0) and F₂(-c, 0) be the foci, using the distance formula, we can find that the sum of the distances from P to the foci is equal to √((x-c)² + y²) + √((x+c)² + y²). Since this sum should be constant, we can conclude that (x-c)² + y² = (x+c)² + y². Expanding and simplifying, we get x² − y² = a² − b².