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Prove that the Curves Y2 = 4x and X2 + Y2 - 6x + 1 = 0 Touch Each Other at the Point (1, 2) ?

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प्रश्न

Prove that the curves y2 = 4x and x2 + y2 - 6x + 1 = 0 touch each other at the point (1, 2) ?

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उत्तर

\[\text { Given }: \]

\[ y^2 = 4x . . . . . \left( 1 \right) \text { and }\]

\[ x^2 + y^2 - 6x + 1 = 0 . . . . . \left( 2 \right)\]

\[\text { From} \left( 1 \right) and \left( 2 \right), \text { we get }\]

\[ x^2 + 4x - 6x + 1 = 0\]

\[ \Rightarrow x^2 - 2x + 1 = 0\]

\[ \Rightarrow \left( x - 1 \right)^2 = 0\]

\[ \Rightarrow x - 1 = 0\]

\[ \Rightarrow x = 1\]

\[\text { Substititing } x = 1 in \left( 1 \right), \text { we get }\]

\[ y^2 = 4\]

\[ \Rightarrow y = \pm 2\]

\[\text { So, the two given curves touch each other at two points} \left( 1, 2 \right) \text { and } \left( 1, - 2 \right) .\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Tangents and Normals - Exercise 16.3 [पृष्ठ ४०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 15 Tangents and Normals
Exercise 16.3 | Q 7 | पृष्ठ ४०
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