![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-12-maths-chapter-16-tangents-and-normals-image-1.gif\" "\\"RD")
<\/figure>\n\n\n\nSolution:<\/strong><\/p>\n\n\n\n (ii) y = \u221ax at x = 9<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n \u21d2 The Slope of the normal = \u2013 6<\/p>\n\n\n\n (iii) y = x3<\/sup> \u2013 x at x = 2<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (iv) y = 2x2<\/sup> + 3 sin x at x = 0<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (v) x = a (\u03b8 \u2013 sin \u03b8), y = a (1 + cos \u03b8) at \u03b8 = -\u03c0 \/2<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (vi) x = a cos3<\/sup> \u03b8, y = a sin3<\/sup> \u03b8 at \u03b8 = \u03c0 \/4<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (vii) x = a (\u03b8 \u2013 sin \u03b8), y = a (1 \u2013 cos \u03b8) at \u03b8 = \u03c0 \/2<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (viii) y = (sin 2x + cot x + 2)2<\/sup> at x = \u03c0 \/2<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (ix) x2<\/sup> + 3y + y2<\/sup> = 5 at (1, 1)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (x) x y = 6 at (1, 6)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n 2. Find the values of a and b if the Slope of the tangent to the curve x y + a x + by = 2 at (1, 1) is 2.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n \u21d2 \u2013 a \u2013 1 = 2(1 + b)<\/p>\n\n\n\n \u21d2 \u2013 a \u2013 1 = 2 + 2b<\/p>\n\n\n\n \u21d2 a + 2b = \u2013 3 \u2026 (1)<\/p>\n\n\n\n Also, the point (1, 1) lies on the curve xy + ax + by = 2, we have<\/p>\n\n\n\n 1 \u00d7 1 + a \u00d7 1 + b \u00d7 1 = 2<\/p>\n\n\n\n \u21d2 1 + a + b = 2<\/p>\n\n\n\n \u21d2 a + b = 1 \u2026 (2)<\/p>\n\n\n\n From (1) & (2), we get b = -4<\/p>\n\n\n\n Substitute b = \u2013 4 in a + b = 1<\/p>\n\n\n\n a \u2013 4 = 1<\/p>\n\n\n\n \u21d2 a = 5<\/p>\n\n\n\n So the value of a = 5 & b = \u2013 4<\/p>\n\n\n\n 3. If the tangent to the curve y = x3<\/sup> + a x + b at (1, \u2013 6) is parallel to the line x \u2013 y + 5 = 0, find a and b<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n The given line is x \u2013 y + 5 = 0<\/p>\n\n\n\n y = x + 5 is the form of equation of a straight line y = mx + c, where m is the Slope of the line.<\/p>\n\n\n\n So the slope of the line is y = 1 \u00d7 x + 5<\/p>\n\n\n\n So the Slope is 1. \u2026 (2)<\/p>\n\n\n\n Also the point (1, \u2013 6) lie on the tangent, so<\/p>\n\n\n\n x = 1 & y = \u2013 6 satisfies the equation, y = x3<\/sup> + ax + b<\/p>\n\n\n\n \u2013 6 = 13<\/sup> + a \u00d7 1 + b<\/p>\n\n\n\n \u21d2 \u2013 6 = 1 + a + b<\/p>\n\n\n\n \u21d2 a + b = \u2013 7 \u2026 (3)<\/p>\n\n\n\n Since, the tangent is parallel to the line, from (1) & (2)<\/p>\n\n\n\n Hence, 3 + a = 1<\/p>\n\n\n\n \u21d2 a = \u2013 2<\/p>\n\n\n\n From (3)<\/p>\n\n\n\n a + b = \u2013 7<\/p>\n\n\n\n \u21d2 \u2013 2 + b = \u2013 7<\/p>\n\n\n\n \u21d2 b = \u2013 5<\/p>\n\n\n\n So the value is a = \u2013 2 & b = \u2013 5<\/p>\n\n\n\n 4. Find a point on the curve y = x3<\/sup> \u2013 3x where the tangent is parallel to the chord joining (1, \u2013 2) and (2, 2).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n 5. Find a point on the curve y = x3<\/sup> \u2013 2x2<\/sup> \u2013 2x at which the tangent lines are parallel to the line y = 2x \u2013 3.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given the curve y = x3<\/sup> \u2013 2x2<\/sup> \u2013 2x and a line y = 2x \u2013 3<\/p>\n\n\n\n First, we will find the slope of tangent<\/p>\n\n\n\n y = x3<\/sup> \u2013 2x2<\/sup> \u2013 2x<\/p>\n\n\n\n y = 2x \u2013 3 is the form of equation of a straight line y = mx + c, where m is the Slope of the line.<\/p>\n\n\n\n So the slope of the line is y = 2 \u00d7 (x) \u2013 3<\/p>\n\n\n\n Thus, the Slope = 2. \u2026 (2)<\/p>\n\n\n\n From (1) & (2)<\/p>\n\n\n\n \u21d2 3x2<\/sup> \u2013 4x \u2013 2 = 2<\/p>\n\n\n\n \u21d2 3x2<\/sup> \u2013 4x = 4<\/p>\n\n\n\n \u21d2 3x2<\/sup> \u2013 4x \u2013 4 = 0<\/p>\n\n\n\n We will use factorization method to solve the above Quadratic equation.<\/p>\n\n\n\n \u21d2 3x2<\/sup> \u2013 6x + 2x \u2013 4 = 0<\/p>\n\n\n\n \u21d2 3 x (x \u2013 2) + 2 (x \u2013 2) = 0<\/p>\n\n\n\n \u21d2 (x \u2013 2) (3x + 2) = 0<\/p>\n\n\n\n \u21d2 (x \u2013 2) = 0 & (3x + 2) = 0<\/p>\n\n\n\n \u21d2 x = 2 or<\/p>\n\n\n\n x = -2\/3<\/p>\n\n\n\n Substitute x = 2 & x = -2\/3 in y = x3<\/sup> \u2013 2x2<\/sup> \u2013 2x<\/p>\n\n\n\n When x = 2<\/p>\n\n\n\n \u21d2 y = (2)3<\/sup> \u2013 2 \u00d7 (2)2<\/sup> \u2013 2 \u00d7 (2)<\/p>\n\n\n\n \u21d2 y = 8 \u2013 (2 \u00d7 4) \u2013 4<\/p>\n\n\n\n \u21d2 y = 8 \u2013 8 \u2013 4<\/p>\n\n\n\n \u21d2 y = \u2013 4<\/p>\n\n\n\n 6. Find a point on the curve y2<\/sup> = 2x3<\/sup> at which the Slope of the tangent is 3<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given the curve y2<\/sup> = 2x3<\/sup> and the Slope of tangent is 3<\/p>\n\n\n\n y2<\/sup> = 2x3<\/sup><\/p>\n\n\n\n Differentiating the above with respect to x<\/p>\n\n\n\n dy\/dx = 0 which is not possible.<\/p>\n\n\n\n So we take x = 2 and substitute it in y2<\/sup> = 2x3<\/sup>, we get<\/p>\n\n\n\n y2<\/sup> = 2(2)3<\/sup><\/p>\n\n\n\n y2<\/sup> = 2 \u00d7 8<\/p>\n\n\n\n y2<\/sup> = 16<\/p>\n\n\n\n y = 4<\/p>\n\n\n\n Thus, the required point is (2, 4)<\/p>\n\n\n\n 7. Find a point on the curve x y + 4 = 0 at which the tangents are inclined at an angle of 45o<\/sup> with the x\u2013axis.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Substitute in xy + 4 = 0, we get<\/p>\n\n\n\n \u21d2 x (\u2013 x) + 4 = 0<\/p>\n\n\n\n \u21d2 \u2013 x2<\/sup> + 4 = 0<\/p>\n\n\n\n \u21d2 x2<\/sup> = 4<\/p>\n\n\n\n \u21d2 x = So when x = 2, y = \u2013 2<\/p>\n\n\n\n And when x = \u2013 2, y = 2<\/p>\n\n\n\n Thus, the points are (2, \u2013 2) & (\u2013 2, 2)<\/p>\n\n\n\n 8. Find a point on the curve y = x2<\/sup> where the Slope of the tangent is equal to the x \u2013 coordinate of the point.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given the curve is y = x2<\/sup><\/p>\n\n\n\n y = x2<\/sup><\/p>\n\n\n\n Differentiating the above with respect to x<\/p>\n\n\n\n From (1) & (2), we get,<\/p>\n\n\n\n 2x = x<\/p>\n\n\n\n \u21d2 x = 0.<\/p>\n\n\n\n Substituting this in y = x2<\/sup>, we get,<\/p>\n\n\n\n y = 02<\/sup><\/p>\n\n\n\n \u21d2 y = 0<\/p>\n\n\n\n Thus, the required point is (0, 0)<\/p>\n\n\n\n 9. At what point on the circle x2<\/sup> + y2<\/sup> \u2013 2x \u2013 4y + 1 = 0, the tangent is parallel to x \u2013 axis.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n \u21d2 \u2013 (x \u2013 1) = 0<\/p>\n\n\n\n \u21d2 x = 1<\/p>\n\n\n\n Substituting x = 1 in x2<\/sup> + y2<\/sup> \u2013 2x \u2013 4y + 1 = 0, we get,<\/p>\n\n\n\n \u21d2 12<\/sup> + y2<\/sup> \u2013 2(1) \u2013 4y + 1 = 0<\/p>\n\n\n\n \u21d2 1 \u2013 y2<\/sup> \u2013 2 \u2013 4y + 1 = 0<\/p>\n\n\n\n \u21d2 y2<\/sup> \u2013 4y = 0<\/p>\n\n\n\n \u21d2 y (y \u2013 4) = 0<\/p>\n\n\n\n \u21d2 y = 0 and y = 4<\/p>\n\n\n\n Thus, the required point is (1, 0) and (1, 4)<\/p>\n\n\n\n 10. At what point of the curve y = x2<\/sup> does the tangent make an angle of 45o<\/sup> with the x\u2013axis?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given the curve is y = x2<\/sup><\/p>\n\n\n\n Differentiating the above with respect to x<\/p>\n\n\n\n \u21d2 y = x2<\/sup><\/p>\n\n\n\n Exercise 16.2 Page No: 15.27<\/p>\n\n\n\n 1. Find the equation of the tangent to the curve \u221ax + \u221ay = a, at the point (a2<\/sup>\/4, a2<\/sup>\/4).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n 2. Find the equation of the normal to y = 2x3<\/sup> \u2013 x2<\/sup> + 3 at (1, 4).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n 3. Find the equation of the tangent and the normal to the following curves at the indicated points:<\/strong><\/p>\n\n\n\n (i) y = x4<\/sup> \u2013 6x3<\/sup> + 13x2<\/sup> \u2013 10x + 5 at (0, 5)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (ii) y = x4<\/sup> \u2013 6x3<\/sup> + 13x2<\/sup> \u2013 10x + 5 at x = 1 y = 3<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (iii) y = x2<\/sup> at (0, 0)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given y = x2<\/sup> at (0, 0)<\/p>\n\n\n\n By differentiating the given curve, we get the slope of the tangent<\/p>\n\n\n\n m (tangent) at (x = 0) = 0<\/p>\n\n\n\n Normal is perpendicular to tangent so, m1<\/sub>m![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-12-maths-chapter-16-tangents-and-normals-image-2.png\" "\\"RD")
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