Question

Let the two foci of an ellipse be (-1,0) and (3, 4) and the foot of perpendicular from the focus (3, 4) upon a tangent to the ellipse be (4,6).

• A. x² + y² - 2x - 4y - 5 = 0
• B. x² + y² - 2x - 4y - 20 = 0
• C. x² + y² + 2x + 4y - 20 = 0
• D. x² + y² + 2x + 4y - 5 = 0

Answer 1

Answer: (B)

x² + y² - 2x - 4y - 20 = 0

Answer 2

1. Center of the ellipse (C): The center is the midpoint of the foci $F_1(-1, 0)$ and $F_2(3, 4)$. $C = \left(\frac{-1+3}{2}, \frac{0+4}{2}\right) = (1, 2)$.
2. Radius of the auxiliary circle (a): The foot of the perpendicular from a focus to a tangent lies on the auxiliary circle. Thus, $P(4, 6)$ lies on the auxiliary circle centered at $C(1, 2)$. The radius of the auxiliary circle is $a$. $a = CP = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
3. Equation of the auxiliary circle: With center $C(1, 2)$ and radius $a=5$, the equation is: $(x-1)^2 + (y-2)^2 = 5^2$ $x^2 - 2x + 1 + y^2 - 4y + 4 = 25$ $x^2 + y^2 - 2x - 4y - 20 = 0$.