What is the largest value of $a$ $a$ for which the circle $x^{2} + y^{2} = a^{2}$ $x^2+y^2=a^2$ lies completely inside the parabola $y^{2} = 4 (x + 4)$ $y^2=4(x+4)$ ?

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What is the largest value of $a$ $a$ for which the circle $x^{2} + y^{2} = a^{2}$ $x^2+y^2=a^2$ lies completely inside the parabola $y^{2} = 4 (x + 4)$ $y^2=4(x+4)$ ?

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Answer 1 · 2018-01-28T14:46:45

$a = 2 \sqrt{3}$ $a=2\sqrt{3}$ .

Explanation:

Let's work this out. We have

$y^{2} = a^{2} - x^{2}$ $y^2=a^2-x^2$

$y^{2} = 4 (x + 4)$ $y^2=4(x+4)$

So then when the circle intersects the parabola the values of $y^{2}$ $y^2$ must agree:

$a^{2} - x^{2} = 4 (x + 4)$ $a^2-x^2=4(x+4)$

$x^{2} + 4 x + (16 - a^{2}) = 0$ $x^2+4x+(16-a^2)=0$

We now find the positive value of $a$ $a$ where the quadratic equation has just one root, where the circle will just touch the parabola. To get that the discriminant must be zero:

(Middle coefficient)^2-(4×(product of other coefficients))=0

$4^{2} - (4 \times (16 - a^{2})) = 0$ $4^2-(4×(16-a^2))=0$

$a^{2} = 12 = 2^{2} \times 3$ $a^2=12=2^2×3$

$a = 2 \sqrt{3}$ $a=2\sqrt{3}$ .

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What is the largest value of $a$ $a$ for which the circle $x^{2} + y^{2} = a^{2}$ $x^2+y^2=a^2$ lies completely inside the parabola $y^{2} = 4 (x + 4)$ $y^2=4(x+4)$ ?

1 Answer

Explanation:

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What is the largest value of aa for which the circle x2+y2=a2x^2+y^2=a^2 lies completely inside the parabola y2=4(x+4)y^2=4(x+4) ?

1 Answer

Explanation:

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What is the largest value of $a$ $a$ for which the circle $x^{2} + y^{2} = a^{2}$ $x^2+y^2=a^2$ lies completely inside the parabola $y^{2} = 4 (x + 4)$ $y^2=4(x+4)$ ?