The tangents to the curve y = (x - 2) ^ 2 - 1 at its points of intersection with the line x - y = 3, intersect at the point:

Correct answer: B. ((5)/(2), - 1) Let coordinates of C be (h, k) chord of contact of C w. r. t. y = (x - 2) ^ 2 - 1 ⇒ y = x ^ 2 - 4x + 3 is T = 0 (y + k)/(2) = xh + 3 - 2(x + h) ⇒ 2(h - 2)x - y = - (6 - 4h - k) compare it with x - y = 3 2(h - 2) = 1 ⇒ h = (5)/(2) 6 - 4h - k = - 3 ⇒ k = - 1 ∴ C is ((5)/(2), - 1)

JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Parabola

The tangents to the curve y = (x - 2) ^ 2 - 1 at its points of...

The tangents to the curve

y = {\left(x - 2\right)}^{2} - 1

at its points of intersection with the line

x - y = 3,

intersect at the point:

Options

A $\left(\dfrac{5}{2}, 1\right)$
B $\left(\dfrac{5}{2}, - 1\right)$
C $\left(- \dfrac{5}{2}, - 1\right)$
D $\left(- \dfrac{5}{2}, 1\right)$

Correct answer

B. $\left(\dfrac{5}{2}, - 1\right)$

Step-by-step solution

Let coordinates of $C$ be \left(h, k\right) $chord of contact of$ C $w. r. t.$ y = {\left(x - 2\right)}^{2} - 1 $$ ⇒ y = {x}^{2} - 4x + 3 $is$ T = 0 $$ \dfrac{y + k}{2} = xh + 3 - 2\left(x + h\right) $$ ⇒ 2\left(h - 2\right)x - y = - \left(6 - 4h - k\right) $compare it with$ x - y = 3 $$ 2\left(h - 2\right) = 1 ⇒ h = \dfrac{5}{2} $$ 6 - 4h - k = - 3 ⇒ k = - 1 $$ ∴ C $is \left(\dfrac{5}{2}, - 1\right)$

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