Ask Questions, Get Answers

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

Find the equation of the two tangents that can be drawn, from the point $(1 , 3 )$ to the ellipse $4x^{2}+9y^{2}=36$

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:
Here $a^2=9,b^2=4$
Let tangent be of the form $y=mx+\sqrt{a^2m^2+b^2}$
(i.e) $y=mx+\sqrt{9m^2+4}$
Step 2:
The tangent passes through $(1,3)$
Therefore $3=m+\sqrt{9m^2+4}$
Therefore $9+m^2-6m=9m^2+4$
$\Rightarrow 8m^2+6m-5=0$
$\Rightarrow \large\frac{-6\pm14}{16}=\large\frac{-20}{16},\large\frac{8}{16}$
(i.e) $\Rightarrow \large\frac{-5}{4},\frac{1}{2}$
Step 3:
The two tangents pass through $(1,3)$.
When $m=\large\frac{1}{2}$
When $m=\large\frac{-5}{4}$
Step 4:
The two tangents are $x-2y+5=0,5x+4y-17=0$
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