{"id":31363,"date":"2024-02-08T15:36:46","date_gmt":"2024-02-08T10:06:46","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=31363"},"modified":"2024-02-09T17:44:45","modified_gmt":"2024-02-09T12:14:45","slug":"ncert-solutions-for-class-12-maths-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-3\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the slope of the tangent to the curve y = 3x4<\/sup> – 4x at x = 4.
\nSolution:
\ny = 3x4<\/sup> – 4x
\nDifferentiating w.r.t. x,
\n\u2234 \\(\\frac { dy }{ dx }\\) = 12x3<\/sup> – 4
\n\u2234 Req. slope = \\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=4 }\\)
\n= 12(43<\/sup>) – 4 = 764.<\/p>\n

Question 2.
\nFind the slope of the tangent to the curve y = \\(\\frac { x-1 }{ x-2 }\\), x \u2260 2 at x = 10.
\nSolution:
\ny = \\(\\frac { x-1 }{ x-2 }\\), x \u2260 2
\nDifferentiating w.r.t. x,
\n\"NCERT<\/p>\n

Question 3.
\nFind the slope of the tangent to curve y = x3<\/sup> – x + 1 at the point whose x-coordinate is 2.
\nSolution:
\nThe curve is y = x3<\/sup> – x + 1
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 1
\n\u2234 slope of tangent = \\(\\frac { dy }{ dx }\\)
\n= 3x\u00b2 – 1
\n= 3 x 2\u00b2 – 1
\n= 11<\/p>\n

Question 4.
\nFind the slope of the tangent to the curve y = x3<\/sup> – 3x + 2 at the point whose x-coordinate is 3.
\nSolution:
\nThe curve is y = x3<\/sup> – 3x + 2
\n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 3
\n\u2234 slope of tangent = \\(\\frac { dy }{ dx }\\)
\n= 3x\u00b2 – 3
\n= 3 x 3\u00b2 – 3
\n= 24<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nFind the slope of the normal to the curve x = a cos3<\/sup> \u03b8, y = a sin3<\/sup> \u03b8 at \u03b8 = \\(\\frac { \\pi }{ 4 } \\).
\nSolution:
\ny = a sin3<\/sup>\u03b8 and x = a cos3<\/sup>\u03b8
\nDifferentiating w.r.t. x,
\n\"NCERT<\/p>\n

Question 6.
\nFind the slope of the normal to the curve x = 1 – a sin \u03b8, y = b cos\u00b2 \u03b8 at \u03b8 = \\(\\frac { \\pi }{ 2 } \\)
\nSolution:
\nx = 1 – a sin \u03b8 and y = b cos\u00b2 \u03b8
\nDifferentiating w.r.t. x,
\n\"NCERT<\/p>\n

Question 7.
\nFind points at which the tangent to the curve y = x3<\/sup> – 3x2<\/sup> – 9x + 7 is parallel to the x-axis.
\nSolution:
\ny = x3<\/sup> – 3x2<\/sup> – 9x + 7
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 3(x – 3)(x + 1)
\nTangent is parallel to x-axis if the slope of tangent = 0
\nor \\(\\frac { dy }{ dx }=0\\)
\n\u21d2 3(x + 3)(x + 1) = 0
\n\u21d2 x = – 1, 3
\nwhen x = – 1, y = 12 & When x = 3, y = – 20
\nHence the tangent to the given curve are parallel to x-axis at the points (- 1, – 12), (3, – 20)<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nFind a point on the curve y = (x – 2)\u00b2 at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
\nSolution:
\nThe equation of the curve is y = (x – 2)\u00b2
\nDifferentiating w.r.t x
\n\\(\\frac { dy }{ dx }\\) = 2(x – 2 )
\nThe point A and B are (2,0) and (4,4) respectively.
\n\"NCERT
\nSlope of AB = \\(\\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }\\)
\n= \\(\\frac { 4-0 }{ 4-2 } =\\frac { 4 }{ 2 } \\) = 2 … (i)
\nSlope of the tangent = 2 (x – 2) ….(ii)
\nfrom (i) & (ii) 2 (x – 2) = 2
\n\u2234 x – 2 = 1 or x = 3
\nwhen x = 3,y = (3 – 2)\u00b2 = 1
\n\u2234 The tangent is parallel to the chord AB at (3, 1)<\/p>\n

Question 9.
\nFind the point on the curve y = x3<\/sup> – 11x + 5 at which the tangent is y = x – 11.
\nSolution:
\nHere, y = x3<\/sup> – 11x + 5
\n\u21d2 \\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 11
\nThe slope of tangent line y = x – 11 is 1
\n\u2234 3x\u00b2 – 11 = 1
\n\u21d2 3x\u00b2 = 12
\n\u21d2 x\u00b2 = 4, x = \u00b12
\nWhen x = 2, y = – 9 & when x = -2,y = -13
\nBut (-2, -13) does not lie on the curve
\n\u2234 y = x – 11 is the tangent at (2, -9)<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nFind the equation of all lines having slope -1 that are tangents to the curve y = \\(\\frac { 1 }{ x-1 }\\), x \u2260 1
\nSolution:
\ny = \\(\\frac { 1 }{ x-1 }\\), x \u2260 1
\nDifferentiating w.r.t x
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac{-1}{(x-1)^{2}}\\)
\nSince the tangent have slope – 1,
\n\\(\\frac { dy }{ dx }\\) = – 1 \u21d2 \\(\\frac{-1}{(x-1)^{2}}\\) = – 1
\n\u21d2 (x – 1)\u00b2 = 1 \u21d2 x – 1 = \u00b1 1
\n\u21d2 x = 2 or x = 0
\nWhen x = 0, y = \\(\\frac { 1 }{ 0 – 1 }\\) = – 1
\nWhen x = 2, y = \\(\\frac { 1 }{ 2 – 1 }\\) = 1
\n\u2234 The required points are (0, – 1) and (2, 1)
\nEquation of the tangent at (0, – 1) is
\ny – 1 = – 1 (x – 0) \u21d2 x + y + 1 = 0
\nEquation of the tangent at (2, 1) is
\ny – 1 = – 1(x – 2) \u21d2 x + y – 3 = 0<\/p>\n

Question 11.
\nFind the equation of ail lines having slope 2 which are tangents to the curve y = \\(\\frac { 1 }{ x-3 }\\), x \u2260 3.
\nSolution:
\nHere y = \\(\\frac { 1 }{ x-3 }\\)
\n\\(\\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\\frac { -1 }{ { (x-3) }^{ 2 } } \\)
\n\u2235 slope of tangent = 2
\n\\(\\frac { -1 }{ { (x-3) }^{ 2 } } =2\\Rightarrow { (x-3) }^{ 2 }=-\\frac { 1 }{ 2 } \\)
\nWhich is not possible as (x – 3)\u00b2 > 0
\nThus, no tangent to y = \\(\\frac { 1 }{ x-3 }\\)
\nHence there is no tangent to the curve with slope 2.<\/p>\n

Question 12.
\nFind the equations of all lines having slope 0 which are tangent to the curve y = \\(\\frac { 1 }{ { x }^{ 2 }-2x+3 }\\)
\nSolution:
\ny = \\(\\frac { 1 }{ { x }^{ 2 }-2x+3 }\\)
\nDifferentiating w.r.t x
\n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac{-(2 x-2)}{\\left(x^{2}-2 x+3\\right)^{2}}\\)
\nSince the slope of the tangent is zero,
\n\\(\\frac { dy }{ dx }\\) = 0
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nFind points on the curve \\(\\frac { { x }^{ 2 } }{ 9 } + \\frac { { y }^{ 2 } }{ 16 } \\) = 1 at which the tangents are
\ni. parallel to x-axis
\nii. parallel to y-axis
\nSolution:
\nThe equation of the curve is \\(\\frac { { x }^{ 2 } }{ 9 } +\\frac { { y }^{ 2 } }{ 16 }\\) = 1 … (i)
\nDifferentiating both sides w.r.t x
\n\\(\\frac { 2 }{ 9 }\\)x + \\(\\frac { 2y }{ 16 }\\)\\(\\frac { dy }{ dx }\\) = 0
\n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { -16x }{ 9y }\\)<\/p>\n

i. Since the tangent is parallel to x axis,
\n\\(\\frac { dy }{ dx }\\) = 0 \u21d2 \\(\\frac{-16x}{9y}\\)
\nHence x = 0
\nWhen x = 0, we get \\(\\frac{0}{9}\\) + \\(\\frac { { y }^{ 2 } }{ 16 }\\) = 1
\n\u21d2 y\u00b2 = 16 \u21d2 y = \u00b14
\n\u2234 (0,4) and (0, – 4) are the points at which the tangents are parallel to the x axis<\/p>\n

ii. Since the tangent is parallel to y axis, its slope is not defined, then the normal is parallel to x-axis whose slope is zero.
\n\"NCERT
\n\u2234 (3, 0) and (- 3, 0) are the points at which the tangents are parallel to they axis.<\/p>\n

Question 14.
\nFind the equations of the tangent and normal to the given curves at the indicated points:
\n(i) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (0,5)
\n(ii) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (1,3)
\n(iii) y = x3<\/sup> at (1, 1)
\n(iv) y = x2<\/sup> at (0,0)
\n(v) x = cos t, y = sin t at t = \\(\\frac { \\pi }{ 4 } \\)
\nSolution:
\n(i) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5
\nDifferentiating w.r.t x
\n\"NCERT
\nEquation of the tangent at (0, 5) is
\n(y- 5) = – 10(x- 0)
\ni.e., 10x+y-5 = 0
\nEquation of the normal at (0, 5) is
\n(y – 5) = – 10(x – 0)
\ni.e., x – 10y + 50 = 0<\/p>\n

(ii) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 4x3<\/sup> – 18x2<\/sup> + 26x – 10
\n\\(\\frac { dy }{ dx }\\)(1, 3)<\/sub> = 4(1) – 18(1) + 26(1) – 10 = 2
\nAt (1, 3), the slope of the tangent = 2
\nAt (1, 3), the slope of the normal = \\(\\frac { – 1 }{ 2 }\\)
\nEquation of the tangent at (1, 3) is y – 3 = 2( x – 1)
\ni.e., y = 2x + 1
\nEquation of the normal at (1, 3) is
\ny – 3 = y(x – 1)
\ni.e., x + 2y – 7 = 0<\/p>\n

(iii) y = x3<\/sup>
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 3x2<\/sup>
\nAt(1, 1), \\(\\frac { dy }{ dx }\\) = 3 x 1\u00b2 = 3
\nAt (1, 1), the slope of the tangent = 3
\nEquation of the tangent at(1, 1) is
\ny – 1 = 3(x – 1)
\ni.e., y = 3x – 2
\nAt (1, 1), the slope o fthe normal = \\(\\frac { -1 }{ 3 }\\)
\nEquation of the normal at (1, 1) is
\ny – 1 = \\(\\frac { -1 }{ 3 }\\) (x – 1)
\ni.e., x + 3y – 4 = 0<\/p>\n

(iv) y = x2<\/sup>
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 2x
\n[ \\(\\frac { dy }{ dx }\\) ](0, 0)<\/sub> = 0
\nAt (0, 0), the slope of the tangent = 0
\nEquation of the tangent at (0, 0) is
\ny – 0 = 0(x – 0)
\ni.e., y = 0 or the x axis
\nSince the tangent is the x axis, the normal is parallel to y axis.
\n\u2234 The equation of the normal at (0, 0) is x = 0, since the normal passes through (0,0)<\/p>\n

(v) y = sin t and x = cos t
\nDifferentiating w.r.t. t, we get \\(\\frac { dy }{ dt }\\) = cos t
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nFind the equation of the tangent line to the curve y = x2<\/sup> – 2x + 7 which is
\n(i) parallel to the line 2x – y + 9 = 0
\n(ii) perpendicular to the line 5y – 15x = 13.
\nSolution:
\ny = x2<\/sup> – 2x + 7
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 2x – 2
\n\\(\\frac { dy }{ dx }\\) = 2(x – 1) … (1)
\n(i) Slope of the line 2x – y + 9 = 0 is 2
\nSince the tangent to the curve is parallel to the line 2x – y + 9 = 0, we get
\n\\(\\frac { dy }{ dx }\\) = 2
\nFrom (1), we get 2(x – 1) = 2 \u21d2 x = 2
\nWhen x = 2, y = 2\u00b2 – 2 x 2 + 7 = 7
\n\u2234 At (2, 7) the tangent is parallel to 2x – y + 9 = 0
\nThe equation of the tangent to the given curve at (2, 7) is y – 7 = 2(x – 2)
\ni.e., 2x – y + 3 = 0 or y – 2x – 3 = 0
\nSlope of the line – 15x = 13 is 3<\/p>\n

(ii) Since the tangent to the curve is perpendicular to the line 5y – 15x = 13,
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac { -1 }{ 3 }\\)
\nFrom (1), we get 2(x – 1) = \\(\\frac { -1 }{ 3 }\\)
\n\"NCERT
\nThe tangent is perpendicular to the line 5y – 15x = 13
\nThe equation of the tangent to the given
\n\"NCERT<\/p>\n

Question 16.
\nShow that the tangents to the curve y = 7x3<\/sup> + 11 at the points where x = 2 and x = – 2 are parallel.
\nSolution:
\nHere, y = 7x3<\/sup> + 11
\n\u21d2 x \\(\\frac { dy }{ dx }\\) = 21 x\u00b2
\nNow m1<\/sub> = slope at x = 2 is
\n\\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=2 }\\)
\n= 21 x 2\u00b2 = 84
\nand m2<\/sub> = slope at x = -2 is
\n\\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=-2 }\\)
\n= 21 x (- 2)\u00b2 = 84
\nHence, m1<\/sub> = m2<\/sub> Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nFind the points on the curve y = x\u00b3 at which the slope of the tangent is equal to the y-coordinate of the point
\nSolution:
\ny = x\u00b3
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 3x\u00b2
\nThe slope of the tangent = 3x\u00b2
\nSince slope of the tangent = y coordinate,
\n3x\u00b2 = y or 3x\u00b2 = x\u00b3
\n= 3x\u00b2 – x\u00b3 = 0 = x\u00b2(3 – x) = 0
\n\u21d2 x = 3 or x = 0
\nWhen x = 0, y = (0)\u00b3 = O
\nWhen x = 3, y= (3)\u00b3 = 27
\n\u2234 (0, 0) and (3, 27) are the points at which the slope of the tangent is equal to the y coordinate of the point.<\/p>\n

Question 18.
\nFor the curve y = 4x3<\/sup> – 2x5<\/sup>, find all the points at which the tangent passes through the origin.
\nSolution:
\nLet, (h, k) be the point at which the tangent passes through the origin.
\ny = 4x\u00b3 – 2x5<\/sup>
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 12x\u00b2 – 10x4<\/sup>
\nAt (h, k) the slope of the tangent = \\(\\frac { dy }{ dx }\\)(h,k)<\/sub> = 12h\u00b2 – 10h4<\/sup>
\nEquation of the tangent at (h, k) is
\ny – k = (12h\u00b2 – 10h4<\/sup>) (x – h)
\nSince the tangent passes through the origin,
\n(0 – k) = (12h\u00b2 – 10h4<\/sup>) (0 – h)
\n\u21d2 k = 12h\u00b3 – 10h5<\/sup> … (1)
\n(h, k) is a point on the curve y = 4x\u00b3 – 2x5<\/sup>
\n\u2234 k = 4h\u00b3 – 2h5<\/sup> … (2)
\nFrom (1) & (2) we get
\n12h\u00b3 – 10h5<\/sup> = 4h\u00b3 – 2h5<\/sup>
\n\u21d2 8h\u00b3 – 8h5<\/sup> = 0
\n\u21d2 8h\u00b3 (1 – h\u00b2) = 0
\n\u21d2 8h\u00b3 (1 – h) (1 +h) = 0
\n\u21d2 h = 0 or h= 1 or h = – 1
\nWhen h = 0, k = 0
\nWhen h = 1, k = 2
\nWhen h = – 1, k = – 2
\n\u2234 (0, 0), (1, 2) and (- 1, – 2) are the points at which the tangent passes through the origin.<\/p>\n

\"NCERT<\/p>\n

Question 19.
\nFind the points on the curve x2<\/sup>\u00a0+ y2<\/sup> – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
\nSolution:
\nHere, x2<\/sup> + y2<\/sup> – 2x – 3 = 0
\n\u21d2 \\(\\frac { dy }{ dx } =\\frac { 1-x }{ y }\\)
\nTangent is parallel to x-axis, if \\(\\frac { dy }{ dx }\\) = 0 i.e.
\nif 1 – x = 0
\n\u21d2 x = 1
\nPutting x = 1 in (i)
\n\u21d2 y = \u00b12
\nHence, the required points are (1,2), (1, -2) i.e. (1, \u00b12).<\/p>\n

Question 20.
\nFind the equation of the normal at the point (am\u00b2, am\u00b3 ) for the curve ay\u00b2 = x\u00b3 .
\nSolution:
\nay\u00b2 = x\u00b3
\nDifferentiating w.r.t. x,
\n\"NCERT<\/p>\n

Question 21.
\nFind the equation of the normal’s to the curve y = x3<\/sup> + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
\nSolution:
\ny = x3<\/sup> + 2x + 6
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 + 2
\nSlope of the normal = \\(\\frac{-1}{3 x^{2}+2}\\)
\nSlope of the line x + 14y + 4 = 0 is \\(\\frac { -1 }{ 14 }\\)
\nSince the normal is parallel to x + 14y + 4 = 0
\n\\(\\frac{-1}{3 x^{2}+2}\\) = \\(\\frac { -1 }{ 14 }\\) \u21d2 3x\u00b2 + 2 = 14
\n\u21d2 3x\u00b2 = 12 \u21d2 x\u00b2 = 4 \u21d2 x = \u00b1 2
\nWhen x = 2, y = 18 and when x = – 2, y = – 6
\n\u2234 At (2, 18) and (- 2, – 6), the normals are parallel to x + 14y + 4 = 0.
\nSlope of the normal = \\(\\frac { -1 }{ 14 }\\)
\nEquation of the normal at (2, 18) is
\ny – 8 = \\(\\frac { -1 }{ 14 }\\)(x – 2)
\ni.e., x + 14y – 254 = 0
\nEquation of the normal at ( -2, -6) is
\ny + 6 = \\(\\frac { -1 }{ 14 }\\) (x + 2) i.e., x + 14y + 86 = 0<\/p>\n

\"NCERT<\/p>\n

Question 22.
\nFind the equations of the tangent and normal to the parabola y\u00b2 = 4ax at the point (at\u00b2,2at).
\nSolution:
\ny\u00b2 = 4ax
\nDifferentiating w.r.t. x,
\n2y\\(\\frac { dy }{ dx }\\) = 4a
\n\"NCERT
\nEquation of the tangent at (at\u00b2, 2at) is
\ny – 2at = \\(\\frac { 1 }{ t }\\)(x – at\u00b2)
\nyt – 2at\u00b2 = x – at\u00b2
\nx – yt + at\u00b2 = 0 or ty = x + at\u00b2
\nSlope of the normal at (at\u00b2, 2at) is – t.
\nEquation of the normal at (at\u00b2, 2at) is
\ny – 2at = – t(x – at\u00b2)
\nxt + y = at\u00b3 + 2at
\ny = – tx + 2at + at\u00b3<\/p>\n

Question 23.
\nProve that the curves x = y\u00b2 and xy = k cut at right angles if 8k\u00b2 = 1.
\nSolution:
\nThe curve are
\nx = y\u00b2 … (1)
\nxy = k … (2)
\nLet P be the point of intersection
\nSubstituting x = y\u00b2 in (2),
\n\"NCERT
\nSince the two curves cut at right angles at P, the product of their slopes at P is – 1.
\n\"NCERT
\nTaking the cubes we get k\u00b2 = \\(\\frac { 1 }{ 8 }\\) or 8k\u00b2 = I<\/p>\n

\"NCERT<\/p>\n

Question 24.
\nFind the equations of the tangent and normal to the hyperbola \\(\\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\\frac { { y }^{ 2 } }{ { b }^{ 2 } }\\) = 1 at the point (x0<\/sub>, y0<\/sub>).
\nSolution:
\n\"NCERT<\/p>\n

Question 25.
\nFind the equation of the tangent to the curve y = \\(\\sqrt { 3x – 2 } \\) which is parallel to the line 4x – 2y + 5 = 0.
\nSolution:
\ny = \\(\\sqrt { 3x – 2 } \\)
\nDifferentiating w.r.t. x,
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac{3}{2 \\sqrt{3 x-2}}\\)
\nSlope of the tangent = \\(\\frac{3}{2 \\sqrt{3 x-2}}\\)
\nSlope of the line 4x – 2y + 5 = 0 is 2
\nSince the tangent is parallel to the line dy
\n4x – 2y + 5 = 0, we get \\(\\frac { dy }{ dx }\\) = 2
\n\"NCERT
\n\u2234 At(\\(\\frac { 41 }{ 48 }\\), \\(\\frac { 3 }{ 4 }\\)), the tangent is parallel to the line 4x – 2y + 5 = 0
\nEquation of the tangent at (\\(\\frac { 41 }{ 48 }\\), \\(\\frac { 3 }{ 4 }\\)) is
\ny – \\(\\frac { 3 }{ 4 }\\) = 2(x – \\(\\frac { 41 }{ 48 }\\)
\n24y – 18 = 48x – 41
\ni.e., 48x – 24y = 23<\/p>\n

\"NCERT<\/p>\n

Question 26.
\nThe slope of the normal to the curve y = 2x\u00b2 + 3 sin x at x = 0 is
\n(a) 3
\n(b) \\(\\frac { 1 }{ 3 }\\)
\n(c) -3
\n(d) \\(-\\frac { 1 }{ 3 }\\)
\nSolution:
\n(d) \u2235 y = 2x\u00b2 + 3sinx
\n\u2234 \\(\\frac { dy }{ dx }\\) = 4x+3cosx
\nat
\nx = 0, \\(\\frac { dy }{ dx }\\) = 3
\n\u2234 slope = 3
\n\u21d2 slope of normal is = \\(\\frac { 1 }{ 3 }\\)<\/p>\n

Question 27.
\nThe line y = x + 1 is a tangent to the curve y\u00b2 = 4x at the point
\n(a) (1,2)
\n(b) (2,1)
\n(c) (1,-2)
\n(d) (-1,2)
\nSolution:
\n(a) The curve is y\u00b2 = 4x,
\n\u2234 \\(\\frac { dy }{ dx } =\\frac { 4 }{ 2y } =\\frac { 2 }{ y } \\)
\nSlope of the given line y = x + 1 is 1
\n\u2234 \\(\\frac { 2 }{ y }\\) = 1
\ny = 2 Putting y = 2 in
\n\u21d2 y\u00b2 = 4x
\n\u21d2 2\u00b2 = 4x
\n\u21d2 x = 1
\n\u2234 The Point of contact is (1, 2)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3 Question 1. Find the slope of the tangent to the curve y = 3×4 – 4x at …<\/p>\n

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