JEE PYQ: Parabola Question 20
Question 20 - 2019 (12 Apr Shift 2)
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:
(1) $\left(\frac{5}{2}, 1\right)$
(2) $\left(-\frac{5}{2}, -1\right)$
(3) $\left(\frac{5}{2}, -1\right)$
(4) $\left(-\frac{5}{2}, 1\right)$
Show Answer
Answer: (3)
Solution
Tangent at $(h, k)$: $\frac{1}{2}(y + k) = (x-2)(h-2) - 1$. Given $x - y = 3$: comparing $\frac{2h-4}{1} = \frac{4h - 6 + k}{1}$. Solving: $h = \frac{5}{2}$, $k = -1$.
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