JEE Advanced 2024 — Paper 2 — MATH-II solution

JEE Advanced2024Paper 2MATH-II

Let $A_{1}, B_{1}, C_{1}$ be three points in the $x y$ -plane. Suppose that the lines $A_{1} C_{1}$ and $B_{1} C_{1}$ are tangents to the curve $y^{2} = 8 x$ at $A_{1}$ and $B_{1}$ , respectively. If $O = (0, 0)$ and $C_{1} = (- 4, 0)$ , then which of the following statements is(are) TRUE?

A
The length of the line segment $O A_{1}$ is $43$
B
The length of the line segment $A_{1} B_{1}$ is 16
C
The orthocentre of the triangle $A_{1} B_{1} C_{1}$ is $(0, 0)$
D
The orthocentre of the triangle $A_{1} B_{1} C_{1}$ is $(1, 0)$

Solution

For the parabola $y^{2} = 8 x$ (so $4 a = 8$ , $a = 2$ ), points are parametrised as $(2 t^{2}, 4 t)$ . Let $A_{1} = (2 t_{1}^{2}, 4 t_{1})$ and $B_{1} = (2 t_{2}^{2}, 4 t_{2})$ .

The chord of contact from an external point $(x_{0}, y_{0})$ to $y^{2} = 8 x$ has equation $y y_{0} = 4 (x + x_{0})$ . The point of intersection of tangents at $A_{1}$ and $B_{1}$ is $(2 t_{1} t_{2}, 2 (t_{1} + t_{2}))$ .

Setting this equal to $C_{1} = (- 4, 0)$ : $2 (t_{1} + t_{2}) = 0$ gives $t_{2} = - t_{1}$ , and $2 t_{1} t_{2} = - 4$ gives $t_{1}^{2} = 2$ , so $t_{1} = 2$ , $t_{2} = - 2$ .

Therefore $A_{1} = (4, 42)$ , $B_{1} = (4, - 42)$ .

(A) $O A_{1} = 16 + 32 = 48 = 43$ . True.

(B) $A_{1} B_{1} = 82 \neq = 16$ . False.

(C, D) The triangle is isosceles with $A_{1}, B_{1}$ symmetric about the x-axis. Altitude from $C_{1}$ is the x-axis itself ( $y = 0$ ). The slope of the tangent at $A_{1}$ is $\frac{1}{t _{1}} = \frac{1}{2}$ , so the altitude from $B_{1}$ (perpendicular to $A_{1} C_{1}$ ) has slope $- 2$ , giving line $y - (- 42) = - 2 (x - 4)$ . Setting $y = 0$ : $42 = - 2 (x - 4)$ , so $x - 4 = - 4$ , $x = 0$ . Orthocentre is $(0, 0)$ . (C) is True, (D) is False.

$∴$ Correct options are (A) and (C).

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