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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find points on the curve\( \large\frac{x^2}{9} + \frac{y^2}{16} = 1\) at which the tangents are $(i)\; parallel\; to\; x - axis$

$\begin{array}{1 1} (0,4) ; (0,-4) \\ (0,3 ) ; (0,-3) \\ (0,2); (0,-2) \\ (0,1);(0,-1) \end{array} $

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1 Answer

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Toolbox:
  • If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at the point $P$.
  • Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given :$\large\frac{x^2}{9}+\frac{y^2}{16}$$=1$-----(1)
Now differentiating w.r.t $x$ we get,
$\large\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}$$=0$
$\Rightarrow \large\frac{dy}{dx}$$=\large\frac{-16}{9}.\frac{x}{y}$
Step 2:
If the tangents are parallel to $x$-axis,then $\large\frac{dy}{dx}$$=0$
$\Rightarrow \large\frac{-16}{9}.\frac{x}{y}$$=0$
Therefore $x=0$
Step 3:
Putting $x=0$ in equ(1)
$0+\large\frac{y^2}{16}$$=1$
$\Rightarrow y=\pm 4$
Therefore the tangents are parallel to $x$-axis at $(0,4)$ and $(0,-4)$.
answered Jul 11, 2013 by sreemathi.v
 

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