JEE Advanced 2018 · Paper 1 · Mathematics Q15

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MathematicsSingle correct+3 / -1Question 51 of 54

Passage

Let $S$ be the circle in the $xy$ -plane defined by the equation $x^2 + y^2 = 4$ .

Let $E_1E_2$ and $F_1F_2$ be the chords of $S$ passing through the point $P_0(1, 1)$ and parallel to the x-axis and the y-axis, respectively. Let $G_1G_2$ be the chord of $S$ passing through $P_0$ and having slope $-1$ . Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$ , the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$ , and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$ . Then, the points $E_3$ , $F_3$ , and $G_3$ lie on the curve

x + y = 4

(x-4)^2 + (y-4)^2 = 16

(x-4)(y-4) = 4

xy = 4

Answer and worked solution

Official answer: A

If the tangents at the ends of a chord meet at $(h,k)$ , the chord is the chord of contact $hx+ky=4$ .

Each of the three chords passes through $P_0(1,1)$ , so $h+k=4$ . Hence $E_3$ , $F_3$ and $G_3$ all lie on $x+y=4$ .

Check: the chord $y=1$ has pole $(0,4)$ , and $0+4=4$ .

(A)

Topics:Chord of contact

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JEE Advanced 2018 · Paper 1, Mathematics question 15