Question

If a tangent to the ellipse x²/a²+y²/b²=1, whose centre is C, meets the major and the minor axes at P and Q respectively then a²/CP²+b²/CQ² is equal to

A. 1
B. -1
C. a+b
D. none of these

Answer 1

To find the value of a²/CP²+b²/CQ² in the given ellipse equation, first determine the points of tangency of the tangent line with the major and minor axes. Then calculate the distances CP and CQ using the distance formula. Finally, substitute the values of a, b, CP, and CQ into the expression a²/CP² + b²/CQ² to get the final answer of 2.

Step-by-step explanation:

To find the value of a²/CP²+b²/CQ² in the given ellipse equation, we need to first determine the points of tangency of the tangent line with the major and minor axes. Let's start by finding the coordinates of points P and Q.

For a point on the major axis, the y-coordinate is 0, so substitute y=0 into the ellipse equation: x²/a² + 0/b² = 1. This simplifies to x²/a² = 1, which gives us x = ±a. Therefore, P has coordinates (-a, 0) and (a, 0).

Similarly, for a point on the minor axis, the x-coordinate is 0, so substitute x=0 into the ellipse equation: 0/a² + y²/b² = 1. This simplifies to y²/b² = 1, which gives us y = ±b. Therefore, Q has coordinates (0, -b) and (0, b).

Now we can calculate the distances CP and CQ using the distance formula. The distance between two points (x₁, y₁) and (x₂, y₂) is given by: distance = sqrt((x₂-x₁)² + (y₂-y₁)²).

Let's calculate CP and CQ:

• For CP, the coordinates of C are (0, 0) and the coordinates of P are (-a, 0) and (a, 0). Therefore, CP = sqrt((-a-0)² + (0-0)²) = sqrt(a²) = a.
• For CQ, the coordinates of C are (0, 0) and the coordinates of Q are (0, -b) and (0, b). Therefore, CQ = sqrt((0-0)² + (-b-0)²) = sqrt(b²) = b.

Finally, substitute the values of a, b, CP, and CQ into the expression a²/CP² + b²/CQ²:

a²/CP² + b²/CQ² = a²/(a²) + b²/(b²) = 1 + 1 = 2.

Therefore, the correct answer is A. 2.