Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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$

Can you answer this question?

1 Answer

0 votes
  • 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 $y=mx\pm\sqrt{a^2m^2+b^2}$
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App