Ex 6.3, 23 - Prove that x = y2, xy = k cut at right angles - Ex 6.3

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.3,23 Prove that the curves 𝑥=𝑦2 and 𝑥𝑦=𝑘 cut at right angles if 8𝑘2=1 We need to show that they cut at right angles Two Curve intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other First we Calculate the point of intersection of Curve (1) & (2) 𝑥=𝑦2 𝑥𝑦=𝑘 Putting Value of 𝑥=𝑦2 in (2) ﷐𝑦﷮2﷯﷐𝑦﷯=𝑘 ﷐𝑦﷮3﷯=𝑘 𝑦=﷐𝑘﷮﷐1﷮3﷯﷯ Putting Value of 𝑦=﷐𝑘﷮﷐1﷮3﷯﷯ in (1) 𝑥=﷐﷐﷐𝑘﷮﷐1﷮3﷯﷯﷯﷮2﷯ 𝑥=﷐𝑘﷮﷐2﷮3﷯﷯ Thus , Point of intersection of Curve is ﷐﷐𝒌﷮﷐𝟐﷮𝟑﷯﷯ ,﷐𝒌﷮﷐𝟏﷮𝟑﷯﷯﷯ We know that Slope of tangent to the Curve is ﷐𝑑𝑦﷮𝑑𝑥﷯ We need to show that Curves cut at right Angle ⇒ tangents of their Curves are perpendicular to each other . ⇒ (Slope of tangent to the Curve 𝑥=﷐𝑦﷮2﷯) × Slope of tangent to the Curve 𝑥𝑦=𝑘) =−1 ﷐1﷮2 ﷐𝑘﷮ ﷐1﷮3﷯﷯﷯ × ﷐−1﷮﷐𝑘﷮ ﷐1﷮3﷯﷯﷯=−1 ﷐1﷮2 ﷐𝑘 ﷮﷐1﷮3﷯﷯ ×﷐𝑘﷮﷐1﷮3﷯﷯﷯=−1 ﷐1﷮2 ﷐𝑘﷮ ﷐1﷮3﷯ + ﷐1﷮3﷯﷯ ﷯=1 ﷐1﷮2 ﷐𝑘﷮ ﷐2﷮3﷯﷯﷯=1 1=2﷐𝑘﷮ ﷐2﷮3﷯﷯ 2﷐𝑘﷮ ﷐2﷮3﷯﷯=1 ﷐𝑘﷮ ﷐2﷮3﷯﷯=﷐1﷮2﷯ ﷐﷐﷐𝑘﷮ ﷐2﷮3﷯﷯﷯﷮3﷯=﷐﷐﷐1﷮2﷯﷯﷮3﷯ ﷐𝑘﷮2﷯=﷐1﷮8﷯ ﷐𝟖𝒌﷮𝟐﷯=𝟏 Hence Proved

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