JEE PYQ: Parabola Question 10
Question 10 - 2020 (06 Sep Shift 2)
The centre of the circle passing through the point $(0, 1)$ and touching the parabola $y = x^2$ at the point $(2, 4)$ is:
(1) $\left(\frac{-53}{10}, \frac{16}{5}\right)$
(2) $\left(\frac{6}{5}, \frac{53}{10}\right)$
(3) $\left(\frac{3}{10}, \frac{16}{5}\right)$
(4) $\left(\frac{-16}{5}, \frac{53}{10}\right)$
Show Answer
Answer: (4)
Solution
Circle passes through $A(0,1)$ and $B(2,4)$. Perpendicular bisector of $AB$: $y - \frac{5}{2} = -\frac{2}{3}(x-1) \Rightarrow 4x + 6y = 19$. Normal to parabola at $(2,4)$: $y - 4 = -\frac{1}{4}(x-2) \Rightarrow x + 4y = 18$. Solving: $x = -\frac{16}{5}$, $y = \frac{53}{10}$.
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