Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola $\dfrac{x^2}{25} - \dfrac{y^2}{16} = -1$.

$\dfrac{x^2}{25}-\dfrac{y^2}{16}=-1$ is equivalent to $\dfrac{y^2}{16}-\dfrac{x^2}{25}=1$, a hyperbola with transverse axis along $Y$. Here 'A$^2$' (under $y^2$) $=16$ and 'B$^2$' (under $x^2$) $=25$, so $A=4,\,B=5$. $e^2=1+\dfrac{B^2}{A^2}=1+\dfrac{25}{16}=\dfrac{41}{16} \Rightarrow e=\dfrac{\sqrt{41}}{4}$. - Transverse axis $=2A=8$; conjugate axis $=2B=10$. - Foci $(0,\pm Ae)=(0,\pm\sqrt{41})$. - Directrices $y=\pm\dfrac{A}{e}=\pm\dfrac{16}{\sqrt{41}}=\pm\dfrac{16\sqrt{41}}{41}$. - Latus rectum $=\dfrac{2B^2}{A}=\dfrac{2(25)}{4}=\dfrac{25}{2}$. > [!ANSWER] Transverse $8$, conjugate $10$, $e=\tfrac{\sqrt{41}}{4}$, foci $(0,\pm\sqrt{41})$, directrices $y=\pm\tfrac{16\sqrt{41}}{41}$, latus rectum $\tfrac{25}{2}$.

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equa

Exercise 7.3 · Q65

\dfrac{x^2}{25} - \dfrac{y^2}{16} = -1

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$\dfrac{x^2}{25}-\dfrac{y^2}{16}=-1$ is equivalent to $\dfrac{y^2}{16}-\dfrac{x^2}{25}=1$ , a hyperbola with transverse axis along $Y$ . Here 'A $^2$ ' (under $y^2$ ) $=16$ and 'B $^2$ ' (under $x^2$ ) $=25$ , so $A=4,\,B=5$ .

$e^2=1+\dfrac{B^2}{A^2}=1+\dfrac{25}{16}=\dfrac{41}{16} \Rightarrow e=\dfrac{\sqrt{41}}{4}$ .

Transverse axis $=2A=8$ ; conjugate axis $=2B=10$ .
Foci $(0,\pm Ae)=(0,\pm\sqrt{41})$ .
Directrices $y=\pm\dfrac{A}{e}=\pm\dfrac{16}{\sqrt{41}}=\pm\dfrac{16\sqrt{41}}{41}$ .
Latus rectum $=\dfrac{2B^2}{A}=\dfrac{2(25)}{4}=\dfrac{25}{2}$ .

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Q.Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola $\dfrac{x^2}{25} - \dfrac{y^2}{16} = -1$ .

Concept understanding — Focal Properties and Latus Rectum of a Hyperbola

Q.Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola x225−y216=−1\dfrac{x^2}{25} - \dfrac{y^2}{16} = -125x2​−16y2​=−1.

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Q.Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola $\dfrac{x^2}{25} - \dfrac{y^2}{16} = -1$ .

Unlock everything free for 14 days