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Find points on the curve\( \large\frac{x^2}{9} + \frac{y^2}{16} = 1\) at which the tangents are $(ii)\; parallel\; to\; y - axis$

This is second part of the multi-part question q13

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  • 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}$$=1$-----(1)
Now differentiating w.r.t $x$ we get,
$\Rightarrow \large\frac{dy}{dx}$$=\large\frac{-16}{9}.\frac{x}{y}$
Step 2:
If the tangents are parallel to $y$-axis,then $\large\frac{dx}{dy}$$=0$
Therefore $y=0$
Step 3:
Putting $y=0$ in equ(1)
$\Rightarrow x=\pm 3$
Therefore the tangents are parallel to $y$-axis at $(3,0)$ and $(-3,0))$.
answered Jul 11, 2013 by sreemathi.v

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