Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
The co-ordinates $(2, 3)$ and $(1, 5)$ are the foci of an ellipse which passes through the origin, then the equation of
Tangent at the origin is $(3\sqrt{2} - 5)x + (1 - 2\sqrt{2})y = 0$
Tangent at the origin is $(3\sqrt{2} + 5)x - (1 + 2\sqrt{2})y = 0$
Normal at the origin is $(3\sqrt{2} + 5)x - (2\sqrt{2} + 1)y = 0$
Normal at the origin is $(3\sqrt{2} - 5)x + (1 - 2\sqrt{2})y = 0$
Step-by-Step Solution
Key Concept: The tangent at any point on an ellipse bisects the angle between the focal radii at that point (reflection property), while the normal bisects the external angle. Using the angle bisector formula with slopes of SP and SP' from foci S(2,3) and S'(1,5) to origin P(0,0), we find the tangent and normal equations.
Given $SP'$ has equation $y = 3x/2$ and $S'P$ has equation $y = 5x$, the bisectors of angle $∠SPS'$ are found using the angle bisector formula: $\frac{3x-2y}{\sqrt{13}} = ± \frac{5x-y}{\sqrt{26}}$. This yields two lines: $(3\sqrt{2}-5)x+(1-2\sqrt{2})y = 0$ (tangent) and $(3\sqrt{2}+5)x-(2\sqrt{2}+1)y = 0$ (normal). Points $(2,3)$ and $(1,5)$ lie on opposite sides of the normal equation, confirming it as the normal.
Correct Answer: 1,3