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Show that the Following Curve Intersect Orthogonally at the Indicated Point Y2 = 8x and 2x2 + Y2 = 10 at ( 1 , 2 √ 2 ) ? - Mathematics

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प्रश्न

Show that the following curve intersect orthogonally at the indicated point y2 = 8x and 2x2 +  y2 = 10 at  \[\left( 1, 2\sqrt{2} \right)\] ?

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उत्तर

\[y^2 = 8x . . . \left( 1 \right)\]

\[2 x^2 + y^2 = 10 . . . \left( 2 \right)\]

\[\text { Given point is }\left( 1, 2\sqrt{2} \right)\]

\[\text { Differentiating (1) w.r.t.x,}\]

\[2y\frac{dy}{dx} = 8\]

\[ \Rightarrow \frac{dy}{dx} = \frac{4}{y}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 1, 2\sqrt{2} \right) = \frac{4}{2\sqrt{2}} = \sqrt{2}\]

\[\text { Differentiating (2) w.r.t.x,}\]

\[4x + 2y\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{y}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 1, 2\sqrt{2} \right) = \frac{- 2}{2\sqrt{2}} = \frac{- 1}{\sqrt{2}}\]

\[\text { Since,} m_1 \times m_2 = - 1\]

Hence,  the given curves intersect orthogonally at the given point.

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अध्याय 16: Tangents and Normals - Exercise 16.3 [पृष्ठ ४०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.3 | Q 3.3 | पृष्ठ ४०

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