Ufumbuzi - Mali za ellipses

$$h$$ inawakilisha kuezesha kutoka kwa asili.
$$k$$ inawakilisha y-ezesha kutoka kwa asili.
Kupata thamani za $$h$$ na $$k$$, tumia fomu ya standard ya ellipse ya usawa:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{x^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$
$$h=0$$
$$k=0$$
Kitovu: $$\left(0,0\right)$$

$$a$$ inawakilisha radius mrefu zaidi ya ellipse, ambayo ni sawa na nusu ya mhimili kuu. Hii inaitwa semi-major axis.
Kupata thamani ya $$a$$, tumia fomu ya standard ya ellipse ya usawa:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{x^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$
$$a^2=\frac{41}{4}$$
Chukua mzizi wa mraba wa pande zote mbili za equation:
$$a=3.202$$

Kwa sababu $$a$$ inawakilisha umbali, ina thamani ya chanya tu.

Katika ellipse ya usawa, mhimili kuu unakimbia sambamba na mhimili wa x na kupita kupitia vertisi za ellipse. Pata vertisi kwa kujumlisha na kutoa $$a$$ kutoka kwa x-kioo $$\left(h\right)$$ cha kitovu.

Ili kupata vertex_1, ongeza $$a$$ kwenye coordinate ya x $$\left(h\right)$$ ya kitovu:
Vertex_1: $$\left(h+a,k\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$a=3.202$$
Vertex_1: $$\left(0+3.202,0\right)$$
Vertex_1: $$\left(3.202,0\right)$$

Ili kupata vertex_2, toa $$a$$ kutoka kwenye coordinate ya x ($$h$$) ya kitovu:
Vertex_2: $$\left(h-a,k\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$a=3.202$$
Vertex_2: $$\left(0-3.202,0\right)$$
Vertex_2: $$\left(-3.202,0\right)$$

$$b$$ inawakilisha radius fupi ya elipso, ambayo ni sawa na nusu ya mhimili mdogo. Hii inaitwa nusu-mhimili mdogo.
Kupata thamani ya $$b$$, tumia fomula ya standard ya elipso ya usawa:
$$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

$$\frac{x^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$
$$b^2=\frac{41}{25}$$
Chukua mzizi wa mraba wa pande zote za equation:
$$b=1.281$$
Kwa sababu b inawakilisha umbali, ina thamani ya positive pekee.

Katika elipso ya horizontal, mhimili mdogo unatembea kwa usawa na mhimili wa y na kupita katika co-vertises ya elipso.
Pata co-vertises kwa kujumlisha na kutoa $$b$$ kutoka coordinate ya y $$\left(k\right)$$ ya kitovu.

Ili kupata co-vertex_1, ongeza $$b$$ kwenye coordinate ya y $$\left(k\right)$$ ya kitovu:
Co-vertex_1: $$\left(h,k+b\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$b=1.281$$
Co-vertex_1: $$\left(0,0+1.281\right)$$
Co-vertex_1: $$\left(0,1.281\right)$$

Ili kupata co-vertex_2, toa $$b$$ kutoka coordinate ya y $$\left(k\right)$$ ya kitovu:
Co-vertex_2: $$\left(h,k-b\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$b=1.281$$
Co-vertex_2: $$\left(0,0-1.281\right)$$
Co-vertex_2: $$\left(0,-1.281\right)$$

Urefu wa fokasi ni umbali kutoka katikati ya elipsi hadi kila pointi ya fokasi na kawaida huwakilishwa na $$f$$.

Kupata $$f$$, tumia formula:
$$f=\sqrt{a^2-b^2}$$
$$a^2=\frac{41}{4}$$
$$b^2=\frac{41}{25}$$
Weka $$a^2$$ na $$b^2$$ kwenye formula na urahisishe:

$$f=\sqrt{\frac{41}{4}-\frac{41}{25}}$$

$$f=\sqrt{\frac{861}{100}}$$

$$f=2.934$$

Kwa sababu $$f$$ inawakilisha umbali, ina thamani chanya tu.

Katika duara refu kwa upana, mhimili mkuu unaendesha sawa na mhimili wa x na kupitia foci.
Pata foci kwa kuongeza na kutoa $$f$$ kutoka kwa kikundi cha x $$\left(h\right)$$ cha kitovu.

Ili kupata focus_1, ongeza $$f$$ kwenye x-coordinate $$\left(h\right)$$ ya kitovu:
Focus_1: $$\left(h+f,k\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$f=2.934$$
Focus_1: $$\left(0+2.934,0\right)$$
Focus_1: $$\left(2.934,0\right)$$

Ili kupata focus_2, toa $$f$$ kutoka kwenye x-coordinate $$\left(h\right)$$ ya kitovu:
Focus_2: $$\left(h-f,k\right)$$
Kitovu: $$\left(h,k\right)=\left(0,0\right)$$
$$h=0$$
$$k=0$$
$$f=2.934$$
Focus_2: $$\left(0-2.934,0\right)$$
Focus_2: $$\left(-2.934,0\right)$$

Tumia formula ya eneo la duara refu kupata eneo la duara refu:
$$\pi ·a·b$$
$$a=3.202$$
$$b=1.281$$
Weka $$a$$ na $$b$$ kwenye formula na rahisisha:

$$\pi ·3.202·1.281$$

$$\pi ·4.102$$

Eneo ni sawa na $$4.102\pi$$

Ili kupata kuingilio la x-, weka $$0$$ kwa $$y$$ kwenye equation standard ya elipsi na utatue equation ya pili ya daraja kwa $$x$$.
Bonyeza hapa kwa maelezo ya hatua kwa hatua ya equation ya pili ya daraja.

$$\frac{x^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$

$$\frac{x^2}{\frac{41}{4}}+\frac{0^2}{\frac{41}{25}}=1$$

$$x_1=3.202$$

$$x_2=-3.202$$

Ili kupata kuingilio la y-, weka $$0$$ kwa $$x$$ kwenye equation standard ya elipsi na utatue equation ya pili ya daraja kwa $$y$$.
Bonyeza hapa kwa maelezo ya hatua kwa hatua ya equation ya pili ya daraja.

$$\frac{x^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$

$$\frac{0^2}{\frac{41}{4}}+\frac{y^2}{\frac{41}{25}}=1$$

$$y_1=1.281$$

$$y_2=-1.281$$

Ili kupata eccentricity tumia formula:
$$\frac{\sqrt{a^2-b^2}}{a}$$
$$a^2=\frac{41}{4}$$
$$b^2=\frac{41}{25}$$
$$a=3.202$$
Wekeza $$a^2$$ , $$b^2$$ na $$a$$ kwenye formula:

$$\frac{\sqrt{\frac{41}{4}-\frac{41}{25}}}{3.202}$$

$$\frac{\sqrt{\frac{861}{100}}}{3.202}$$

$$\frac{2.934}{3.202}$$

$$0.916$$

Ubunifu unalingana na $$0.916$$