JEE PYQ: Parabola Question 15
Question 15 - 2019 (08 Apr Shift 2)
The tangent to the parabola $y^2 = 4x$ at the point where it intersects the circle $x^2 + y^2 = 5$ in the first quadrant, passes through the point:
(1) $\left(-\frac{1}{3}, \frac{4}{3}\right)$
(2) $\left(\frac{1}{4}, \frac{3}{4}\right)$
(3) $\left(\frac{3}{4}, \frac{7}{4}\right)$
(4) $\left(-\frac{1}{4}, \frac{1}{2}\right)$
Show Answer
Answer: (3)
Solution
Intersection of $y^2 = 4x$ and $x^2 + y^2 = 5$: $x^2 + 4x - 5 = 0 \Rightarrow x = 1$, $y = 2$. Tangent at $(1,2)$: $2y = 2(x+1) \Rightarrow y = x + 1$. Check $\left(\frac{3}{4}, \frac{7}{4}\right)$: $\frac{7}{4} = \frac{3}{4} + 1$. Yes.
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