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प्रश्न
Find the points on the curve \[\frac{x^2}{4} + \frac{y^2}{25} = 1\] at which the tangent is parallel to the x-axis ?
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उत्तर
The slope of the x-axis is 0.
Now, let (x1, y1) be the required point.
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence,} \frac{{x_1}^2}{4} + \frac{{y_1}^2}{25} = 1 . . . \left( 1 \right) \]
\[\text { Now }, \frac{x^2}{4} + \frac{y^2}{25} = 1 \]
\[ \therefore \frac{2x}{4} + \frac{2y}{25}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{2y}{25}\frac{dy}{dx} = \frac{- x}{2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 25x}{4y}\]
\[\text { Now }, \]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{- 25 x_1}{4 y_1}\]
\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)=\text { Slope of thex-axis [Given] }\]
\[ \therefore \frac{- 25 x_1}{4 y_1} = 0\]
\[ \Rightarrow x_1 = 0\]
\[\text { Also }, \]
\[0 + \frac{{y_1}^2}{25} = 1 [\text { From eq.} (1)]\]
\[ \Rightarrow {y_1}^2 = 25\]
\[ \Rightarrow y_1 = \pm 5\]
\[\text { Thus, the required points are }\left( 0, 5 \right)\text { and }\left( 0, - 5 \right).\]
