JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q29

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MathematicsInteger+4 / -1Question 29 of 90

A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^2=2x$ and touches the parabola $y=\left(x-\dfrac14\right)^2+\alpha$ , where $\alpha>0$ . Then $(4\alpha-8)^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 63

Circle. Through $(0,0)$ and $\left(\frac12,0\right)$ with radius 2: centre $\left(\frac14,k\right)$ , $k^2=4-\frac1{16}$ , $k=\frac{\sqrt{63}}4$ .

Touching. The parabola's vertex $\left(\frac14,\alpha\right)$ lies on the circle's vertical diameter, and it is narrower than the circle, so it touches at the top: $\alpha=k+2$ .

Value. $(4\alpha-8)^2=(4k)^2=63$ .

Answer: 63

Topics:Equation of a circle Parabola

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JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 29