Conic Sections
Conic Section
star_batch_jee_advanced_2025
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 and normal at a point on a conic are angle bisectors of the focal radii.
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