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Question Number 51614 by peter frank last updated on 29/Dec/18

$$Provethatlengthof$$$$lactusrectumwhen$$$$directrixareparallelto$$$$y-axisis$$$$y_1-y_2=\frac{2b^2}{a}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

$$forellipse\frac{x^2}{a^2}+\frac{y^2}{b^2}=1ditectrixis\parallel toyaxis$$$$focus(c,0)and(-c,0)soeqnofstline\parallel toyaxis$$$$passingthrough(c,0)isx=c$$$$solvex=cand\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{y^2}{b^2}=1-\frac{c^2}{a^2}soy=\pm \frac{b}{a}\sqrt{a^2-c^2}$$$$nowdiztancebetween(c,\frac{b}{a}\sqrt{a^2-c^2})and(c,\frac{-b}{a}\sqrt{a^2-c^2})$$$$is=\frac{2b}{a}\sqrt{a^2-c^2}[forellipse{c}^2=a^2-b^2]$$$$=\frac{2b}{a}\times b=\frac{2b^2}{a}$$$$forhyperbola\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$$$directrixis\parallel toyaxis$$$$focus(c,0)(-c,0)$$$$eqnofstlinepassingthrough(c,0)and\parallel toyaxis$$$$isx=c$$$$solvex=cand\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$$$\frac{y^2}{b^2}=\frac{c^2}{a^2}-1$$$$y=\pm \frac{b}{a}\sqrt{c^2-a^2}$$$$y=\pm \frac{b}{a}\times b[forhyperbola{c}^2=a^2+b^2]$$$$y=\pm \frac{b^2}{a}$$$$solatusrectumis\frac{2b^2}{a}$$$$$$

Commented by peter frank last updated on 29/Dec/18

$$verynicesir....thanks$$

Answered by peter frank last updated on 29/Dec/18

$$\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$$$$x=h+ae(paralleltoy-axis)$$$$\frac{{(h+ae-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$$$$e^2-1=\frac{{(y-k)}^2}{b^2}$$$$y-k=\sqrt{b^2(e^2-1)}$$$$y-k=\pm b\sqrt{(e^2-1)}$$$$y=k\pm b\sqrt{(e^2-1)}$$$$y_1=k-b\sqrt{e^2-1}....\left(i\right)$$$$y_2=k+b\sqrt{e^2-1}....\left(ii\right)$$$$y_2-y_1=-2b\sqrt{e^2-1}$$$$y_1-y_2=2b\sqrt{e^2-1}$$$$from$$$$b^2=a^2(e^2-1)$$$$y_1-y_2=2b\times \frac{b}{a}$$$$y_1-y_2=\frac{2b^2}{a}$$$$$$

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