JEE PYQ: Parabola Question 28
Question 28 - 2019 (12 Jan Shift 1)
The tangent to the curve $y = x^2 - 5x + 5$, parallel to the line $2y = 4x + 1$, also passes through the point:
(1) $\left(\frac{7}{2}, \frac{1}{4}\right)$
(2) $\left(\frac{1}{8}, -7\right)$
(3) $\left(-\frac{1}{8}, 7\right)$
(4) $\left(\frac{1}{4}, \frac{7}{2}\right)$
Show Answer
Answer: (2)
Solution
Slope $= 2$. $y’ = 2x - 5 = 2 \Rightarrow x = \frac{7}{2}$, $y = \frac{49}{4} - \frac{35}{2} + 5 = -\frac{1}{4}$. Tangent: $y + \frac{1}{4} = 2(x - \frac{7}{2}) \Rightarrow y = 2x - \frac{29}{4}$. Check $\left(\frac{1}{8}, -7\right)$: $-7 = \frac{1}{4} - \frac{29}{4} = -7$. Yes.
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