Advertisements
Advertisements
प्रश्न
Find the points on the curve \[\frac{x^2}{9} + \frac{y^2}{16} = 1\] at which the tangent is parallel to x-axis ?
Advertisements
उत्तर
The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence,} \frac{{x_1}^2}{9} + \frac{{y_1}^2}{16} = 1 . . . \left( 1 \right)\]
\[\text { Now,} \]
\[\frac{x^2}{9} + \frac{y^2}{16} = 1 \]
\[ \Rightarrow \frac{2x}{9} + \frac{2y}{16}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{y}{16}\frac{dy}{dx} = \frac{- x}{9}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 16x}{9y}\]
\[\text { Now,} \]
\[\text { Slope of the tangent at }\left( x, y \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{- 16 x_1}{9 y_1}\]
\[\text { Slope of the tangent at }\left( x, y \right)=\text { Slope of the x-axis [Given]}\]
\[ \therefore \frac{- 16 x_1}{9 y_1} = 0\]
\[ \Rightarrow x_1 = 0\]
\[0 + \frac{{y_1}^2}{16} = 1 [\text { From eq.} (1)]\]
\[\text { Also,} \]
\[ {y_1}^2 = 16\]
\[ y_1 = \pm 4\]
\[\text { Thus, the required points are }\left( 0, 4 \right)\text { and }\left( 0, - 4 \right).\]
