Tambayoyi da Tattaunawa game da Tangents zuwa Sassan Conic
Pendahuluan
Sashen mazugi wani lanƙwasa ne da ya samo asali daga haɗuwar jirgin sama da mazugi dipolar. Waɗannan lanƙwasa sun haɗa da da'ira, ellipses, parabolas, da hyperbolas. Wani muhimmin batu wajen fahimtar sassan mazugi shine layin tangent. Tangent zuwa sashin mazugi shine layi wanda ke taɓa lanƙwasa mazugi a wuri ɗaya kawai. Wannan labarin zai tattauna misalai da yawa na matsaloli da tattaunawa game da tangent zuwa sassan mazugi.
Mai kama da Da'ira
Da'ira sashe ne mai siffar mazugi wanda ke da siffa mafi sauƙi da kuma cikakkiyar daidaito. Bari mu fara da misalin matsala game da tangents zuwa da'ira.
Misali Tambaya ta 1
An ba da da'ira mai lissafin \( (x – 2)^2 + (y + 3)^2 = 25 \). Kayyade lissafin layin tangent a wurin \(5, -3)\) akan da'irar.
Tattaunawa
Daidaiton da'ira gabaɗaya shine \( (x – h)^2 + (y – k)^2 = r^2 \), tare da \( (h, k) \) a matsayin tsakiyar da'irar da kuma \( r \) a matsayin radius. A cikin wannan matsalar, tsakiyar da'irar \((h, k)\) shine \((2, -3)\) kuma radius \( r = \sqrt{25} = 5 \).
Ana iya samun layin tangent a wurin \((x_1, y_1)\) akan da'irar ta amfani da dabarar da ke ƙasa:
\[ (x – h)(x_1 – h) + (y – k)(y_1 – k) = r^2 \]
Shigar da ƙimomin da aka sani:
\[ (x - 2) (5 - 2) + (y + 3) (-3 + 3) = 25 \]
\[ (x - 2) (3) + (y + 3) (0) = 25 \]
\[ 3(x – 2) = 25 \]
\[ 3x – 6 = 25 \]
\[ 3x = 31 \]
\[ x = \frac{31}{3} \]
Daidaiton layin tangent shine \(x = \frac{31}{3}\), amma akwai kuskure a cikin wannan hanyar saboda ma'aunin \((5, -3)\) a bayyane yake maki ne akan da'irar. Saboda haka, muna amfani da hanyar gargajiya ta hanyar maye gurbin gangaren layin tangent a wannan takamaiman wurin:
Ma'aunin tangent shine, \((5, -3)\), to, gradient (m) na layin radius shine \(m = \frac{-3 – (-3)}{5 – 2}=0\), inda gradient na layin tangent ya zama ba a fayyace shi ba ga tangent tsaye na baya.
Layin Tangent zuwa Ellipse
Ellipse wani sashe ne mai siffar mazugi wanda ke da gatari biyu na daidaito: babban axis (dogon) da ƙaramin axis (gajere). Ga wasu misalan matsalolin da ake samu da ellipses.
Misali Tambaya ta 2
An ba da ellipse tare da lissafin \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). Kayyade lissafin layin tangent a wurin \((2, \frac{3}{2})\) akan ellipse.
Tattaunawa
Daidaiton layin tangent zuwa ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) a wurin \((x_1, y_1)\) shine:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Da \(a = 4\) da \(b = 3\), maye gurbin ƙimar \(a\), \(b\), da kuma ma'anar \((2, \frac{3}{2})\):
\[ \frac{x(2)}{4^2} + \frac{y(\frac{3}{2})}{3^2} = 1 \]
\[ \frac{2x}{16} + \frac{3y}{6} = 1 \]
\[ \frac{x}{8} + \frac{y}{2} = 1 \]
A ninka dukkan lissafin da 8 domin kawar da sassan:
\[ x + 4y = 8 \]
Don haka, daidaiton layin tangent zuwa ellipse shine \( x + 4y = 8 \).
Layin Tangent zuwa Parabola
Parabola sashe ne mai siffar mazugi mai siffar siffa ɗaya da kuma kusurwa ɗaya. Ga wasu misalan matsalolin da ke tattare da parabolas.
Misali Tambaya ta 3
An ba da parabola tare da lissafin \( y^2 = 4x \). Kayyade lissafin layin tangent a wurin \((1, 2)\) akan parabola.
Tattaunawa
Daidaiton layin tangent zuwa parabola \( y^2 = 4ax \) a wurin \((x_1, y_1)\) shine:
\[ yy_1 = 2a(x + x_1) \]
Daga lissafin parabola \( y^2 = 4x \), mun sami \( 4a = 4 \) don haka \( a = 1 \). Sauya ƙimar \( a \) da ma'anar \((1, 2)\):
\[ 2y = 2(1)(x + 1) \]
\[ 2y = 2x + 2 \]
\[y = x + 1 \]
Don haka, daidaiton layin tangent zuwa parabola shine \( y = x + 1 \).
Layin Tangent zuwa Hyperbola
Hyperbola sashe ne mai siffar conic wanda ke da reshe biyu da kuma asymptotes biyu. Ga wasu misalan matsalolin da ke tattare da hyperbolas.
Misali Tambaya ta 4
An ba da hyperbola tare da lissafin \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \). Kayyade lissafin layin tangent a wurin \((5, 0)\) akan hyperbola.
Tattaunawa
Daidaiton layin tangent zuwa hyperbola \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\) a wurin \((x_1, y_1)\) shine:
\[ \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \]
Da \(a = 5 \) da \(b = 4 \), a maye gurbin ƙimar \(a \), \(b \), da kuma maki \((5, 0)\):
\[ \frac{x(5)}{25} – \frac{y(0)}{16} = 1 \]
\[ \frac{5x}{25} – 0 = 1 \]
\[ \frac{x}{5} = 1 \]
\[x = 5 \]
Don haka, daidaiton layin tangent zuwa hyperbola shine \( x = 5 \).
Kammalawa
Tangents zuwa sassan conic suna taka muhimmiyar rawa a lissafi da aikace-aikace daban-daban na aiki. Fahimtar yadda ake nemo daidaiton tangents zuwa nau'ikan sassan conic daban-daban, kamar da'ira, ellipses, parabolas, da hyperbolas, ƙwarewa ce mai mahimmanci a cikin lissafi da lissafi na nazari. Tare da misalan da tattaunawa a sama, ana fatan masu karatu za su sami fahimtar ra'ayoyi da hanyoyin tantance tangents zuwa sassan conic.
