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Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram:$ 16x^{2}+9y^{2}+32x-36y=92$

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Toolbox:
  • Shifting of the origin from $(0,0)$ to $(h,k)$ then if the new axes form $X-Y$ coordinate system then $x=X+a$,$y=Y+k$
  • The new coordinates of a point $P(x,y)$ become $X=x-a,Y=y-k$
  • Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
  • $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/16_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/16_2_Toolbox1.png
  • Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
  • End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae)$ and $(\pm \large\frac{b^2}{a}$$,-ae)$
  • Directrices $y=\pm \large\frac{a}{e}$.
  • The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
  • General form of standard ellipses with centre $C$ with major axis $2a$,minor axis $2b$ and axes parallel to the coordinate axes.
  • $\large\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$$=1$
  • In this case major axis y=k and minor axis $x=h$.
  • $\large\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}$$=1$
  • In this case major axis $x=h$ and minor axis $y=k$.
Step 1:
$16x^2+9y^2+32x-36y=92$
$16x^2+32x+9y^2-36y=92$
Completing the squares,
$16(x^2+2x+1)+9(y^2-4y+4)=92+16+36$
$16(x+1)^2+9(y-2)^2=144$
The above equation is divided by $144$
$\large\frac{(x+1)^2}{9}+\frac{(y-2)^2}{16}$$=1$
Step 2:
Shifting the origin to $(-1,2)$ by translation of axes.
$X=x+1$
$Y=y-2$
$x=X-1$
$y=Y+2$
$\large\frac{X^2}{9}+\frac{Y^2}{16}$$=1$
$a^2=16,b^2=9$
$\Rightarrow a=4,b=3$
Step 3:
The major axes is $X=0$ (i.e) $x+1=0$
Step 4:
$XY$-axes :
eccentricity $e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{9}{16}}$
$\Rightarrow \large\frac{\sqrt 7}{4}$
Centre : $(0,0)$
Foci : $(0,\pm ae)$
$ae=4\times \large\frac{\sqrt 7}{4}$$=\sqrt 7$
$\Rightarrow (0,\pm \sqrt 7)$
Vertices : $(0,\pm a)$
$\Rightarrow (0,\pm 4)$
Step 5:
$xy$-axes :
Centre : $(-1,2)$
Foci : $(-1,2\pm \sqrt 7)$
Vertices : $(-1,2\pm 4)$
$\Rightarrow (-1,6),(-1,-2)$
answered Jun 18, 2013 by sreemathi.v
edited Jun 20, 2013 by sreemathi.v
 

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