finding focal point (1 Viewer)

Sien · Oct 21, 2015

how do you find the focal point of y=x^2+6x+7
i never really learnt that part of maths properly so....

InteGrand · Oct 21, 2015

Sien said:
how do you find the focal point of y=x^2+6x+7
i never really learnt that part of maths properly so....

$First get it into the form $(x-h)^2 = 4a(y-k)$ or $(x-h)^2 = -4a(y-k)$:$

$Completing the square, we have $y=(x+3)^2 -2$$

$$\Rightarrow (x+3)^2 = y+2 \Longleftrightarrow (x-(-3))^2 = 4\times \frac{1}{4}\left(y-(-2) \right)$.$

$So the vertex is $(-3,-2)$, and the focal length is $\frac{1}{4}$ units. The parabola is clearly upwards-facing. The focal length tells us how far the focus is from the vertex. For an upwards-opening parabola like this, the focus is directly above the vertex. This means the $x$-value of the focus is that of the vertex, i.e. $-3$. And the $y$-value of the focus is the focal length plus the $y$-value of the vertex (as the focus is \textsl{above} the vertex by the focal length amount), i.e. focus has $y=-2+\frac{1}{4}=-1\tfrac{3}{4}$.$

$So the focus is $\left(-3,-1\tfrac{3}{4} \right)$.$

$Even though you didn't ask for it, the directrix of a concave up parabola like this is a horizontal line that is $a$ units \textsl{below} the vertex ($a$ is the focal length). So in this case, the directrix is the horizontal line $y=-2-\frac{1}{4}$, i.e. $y=-2\tfrac{1}{4}$.$

pricey97 · Oct 21, 2015

pricey97 · Oct 21, 2015

Whoops, don't worry, intergrand beat me to it [emoji2]

Sien · Oct 21, 2015

Jesus I'm glad I asked this question, would've lost like 4 marks if I didn't

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finding focal point (1 Viewer)

Sien

将来: NEET

InteGrand

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pricey97

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pricey97

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Sien

将来: NEET

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