Question

For the ellipse \[ \frac{x^2}{36}+\frac{y^2}{25}=1, \] the straight line \[ 2x+y-5=0 \] is

• A. a tangent
• B. a normal
• C. a focal chord
• D. a chord not passing through its foci

Hint:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the line \[ y=mx+c \] is tangent if \[ \boxed{c^2=a^2m^2+b^2.} \] If this condition is not satisfied but the line intersects the ellipse at two points, it is a chord.

Answer 1

The Correct Option is D

Step 1: Identify the ellipse. The ellipse is \[ \frac{x^2}{36}+\frac{y^2}{25}=1. \] Here, \[ a=6,\qquad b=5. \] The foci are \[ (\pm c,0), \] where \[ c=\sqrt{a^2-b^2} =\sqrt{36-25} =\sqrt{11}. \] Thus, the foci are \[ (\pm\sqrt{11},0). \]

Step 2: Check whether the line is a tangent. The line is \[ 2x+y-5=0. \] The condition for tangency to the ellipse is \[ c^2=a^2m^2+b^2, \] where the line is written as \[ y=mx+c. \] Here, \[ y=-2x+5, \] so \[ m=-2,\qquad c=5. \] Now, \[ a^2m^2+b^2 = 36(4)+25 = 169, \] whereas \[ c^2=25. \] Since \[ 25\neq169, \] the line is not a tangent.

Step 3: Check whether it passes through a focus. Substituting \[ (\sqrt{11},0) \] into the line, \[ 2\sqrt{11}-5\neq0. \] Similarly, \[ (-\sqrt{11},0) \] also does not satisfy the equation. Hence, the line does not pass through either focus. Since the line intersects the ellipse at two distinct points, it is a chord.

Step 4: Conclude. Therefore, the given line is \[ \boxed{\text{a chord not passing through its foci}.} \] Hence, the correct option is \(\boxed{(D)}\).