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Find the equation of the tangent and normal to the ellipse $\large\frac{x^{2}}{20}+\frac{y^{2}}{5}$$=1,$ Which are perpendicular to $ x+y+z=0$

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Toolbox:
  • Equation of any tangent to
  • $\quad(i)$ The parabola $y^2=4ax$ is of the form $y=mx+\large\frac{a}{m}$
  • $\quad(ii)$ The ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2+b^2}$
  • $\quad(iii)$ The hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Step 1:
Tangent to $\large\frac{x^2}{20}+\frac{y^2}{5}$$=1$ which is $\perp$ to the line $x+y+z=0$
Compare with $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$,
We have $a^2=20,b^2=5$
Any tangent i of the form $y=mx\pm\sqrt{a^2m^2+b^2}$
(i.e)$y=mx\pm\sqrt{20m^2+5}$
Step 2:
The required tangents are $\perp$ to $x+y+z=0$ whose slope is $-1$.
Therefore slope of the tangent =$1=m$
The equation of the tangents are $y=x\pm\sqrt{20+5}$
(i.e) $y=x\pm \sqrt{25}$
$y=x\pm 5$
answered Jun 21, 2013 by sreemathi.v
 

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