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Tangents and its Equations Test 39

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Tangents and its Equations Test 39
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  • Question 1
    1 / -0
    Find the case of tangents & normal to the curve $$y=x^{4}-6x^{3}+13x^{2}-10x+5$$ at $$(1, 3)$$
    Solution
    $$y=x^4-6x^3+13x^2-10x+5$$
    $$\therefore \dfrac{dy}{dx}=4x^3-18x^2+26x-10$$
    $$\therefore \dfrac{dy}{dx}=4(1)^3-18(1)^2+26(1)-10$$
    $$(x=1, y=3)$$
    $$\therefore \dfrac{dy}{dx}=4-18+26-10=2$$
    $$\therefore$$ Equation of Tangent:-
    $$y-3=2(x-1)$$
    $$\therefore y-3=2x-2$$
    $$\therefore y-2x-1=0$$
    $$\therefore$$ Equation of Normal:-
    $$y-3=\dfrac{-1}{2}(x-1)$$
    $$\therefore 2y-6=-x+1$$
    $$\therefore 2y+x-7=0$$.

  • Question 2
    1 / -0
    Two lines drawn through the point $$A ( 4,0 )$$  divide the area bounded by the curve $$y = \sqrt { 2 } \sin ( \pi x / 4 )$$  and  $$x$$ - axis between the lines $$x = 2$$  and   $$x = 4$$  into three equal parts. Sum of the slopes of the drawn lines is:
    Solution

    Area bounded by $$y=\sqrt{2}\sin\left(\dfrac{\pi x}{4}\right)$$ and $$x-axis$$ between the lines $$x=2$$ and $$x=4,$$
    $$\Delta=\sqrt{2}\displaystyle\int_2^4\sin\dfrac{\pi x}{4}dx$$

       $$=\left[-\dfrac{4\sqrt{2}}{\pi}.\cos\dfrac{\pi x}{4}\right]^4_2$$

       $$=\dfrac{4\sqrt{2}}{\pi}\,unit^2$$ 

    Let the drawn lines are $$L_1:y-m_1(x-4)=0$$ and $$L_2:y-m_2(x-4)=0,$$ meeting the line $$x=2$$ at the points $$A$$ and $$B,$$ respectively.
    Clearly, $$A=(2,-2m_1);\,B=(2,-2m_2)$$             [ Fig ]
    Now,
    $$\Delta_{ACD}=\dfrac{\Delta}{3}\Rightarrow \dfrac{4\sqrt{2}}{3\pi}=\dfrac{1}{2}.2.(-2m_1)$$
    $$\Rightarrow$$  $$m_1=-\dfrac{2\sqrt{2}}{3\pi}$$ 
    Also, $$\Delta_{BCD}=\dfrac{2\Delta}{3}$$
    $$\Rightarrow$$  $$\dfrac{8\sqrt{2}}{3\pi}=\dfrac{1}{2}\times 2-2m_2$$
    $$\Rightarrow$$  $$m_2=\dfrac{-4\sqrt{2}}{3\pi}$$
    $$\Rightarrow$$  Required sum $$=-\dfrac{2\sqrt{2}}{3\pi}+\left(\dfrac{-4\sqrt{2}}{3\pi}\right)$$

                                   $$=\dfrac{-6\sqrt{2}}{3\pi}$$

                                   $$=\dfrac{-2\sqrt{2}}{\pi}$$

  • Question 3
    1 / -0
    The tangent to the curve $$2a^2y=x^3-3ax^2$$ is parallel to the x-axis at the points
    Solution
    Let $$(x_1,y_1)$$ represent the required points.

    The slope of the $$x-$$axis is $$0$$

    Here $$2a^2y=x^3-3ax^2$$, since the points lies on the curve we get

    $$2a^2y_1=x_1^3-3ax_1^2 \quad ...(1)$$

    Consider $$2a^2y=x^3-3ax^2$$

    On differentiating both sides with respect to $$x$$, we get

    $$2a^2\dfrac{dy}{dx}=3x^2-6ax$$

    $$\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2-6ax}{2a^2}$$

    Slope of the tangent at $$(x_1,y_1)=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{3x_1^2-6ax_1}{2a^2}$$

    It is given that slope of the tangent at $$(x_1,y_1)=$$ slope of the $$x-$$axis.

    $$\Rightarrow \dfrac{3x_1^2-6ax_1}{2a^2}=0$$

    $$\Rightarrow 3x_1^2-6ax_1=0$$

    $$\Rightarrow x_1(3x_1-6a)=0$$

    $$\Rightarrow x_1=0$$ or $$x_1=2a$$

    Also, from $$(1)$$

    $$2a^2y_1=0$$ or $$2a^2y_1=8a^3-12a^3$$

    $$\Rightarrow y_1=0$$ or $$y_1=-2a$$

    Thus, the required points are $$(0,0)$$ and $$(2a,-2a)$$


  • Question 4
    1 / -0
    The point on the curve $${x}^{2}+{y}^{2}-2x-3=0$$ at which the tangent in parallel to x-axis is 
    Solution
    $$(1,2)$$ and $$(1,-2)$$ are two points where the tangent is parallel to the x-axis.
    The equation of curve can be reduced to $$(x-1)^2+y^2=4$$
    Therefore the center is $$(1.0)$$ on the x-axis and the radius is $$2$$ units
    Now if we add and subtract $$2$$ to the $$y$$ co-ordinate of the center we get $$(1,2)$$ and $$(1,-2)$$ are two points where tangent is possible to x-axis.
    $$y=2$$ and $$y=-2$$ are the two equations of the tangent to the circle and also parallel to axis and equation of the line$$x=1$$ represent diameter perpendicular to $$x-axis$$
  • Question 5
    1 / -0
    Which of the following lines, is a normal to the parabola $${y}^{2}=16x$$?
    Solution
    $$\begin{array}{l} Accoding\, to\, the\, \, question, \\ Equ\, of\, parabola\, is........ \\ y=mx-a{ m^{ 3 } }-2am\, \, \, \, \, \, \, (where\, \, m=slope,\, \, condition\, of\, normal\, \, is,\, c=-2am-a{ m^{ 3 } }) \\ given, \\ y=(x-11)\cos  \theta -\cos  3\theta  \\ \, \, \, \, =x\cos  \theta -11\cos  \theta -\cos  3\theta \, \, \, \, \, (where\, m=\cos  \theta ) \\ \, \, \, \, c=-11\cos  \theta -\cos  3\theta  \\ \, \, \, =-11\cos  \theta -(4co{ s^{ 3 } }\theta -3\cos  \theta ) \\ \, \, \, =\, -8\cos  \theta -4{ \cos ^{ 3 }  }\theta \, is\, the\, required\, option \\ So,the\, correct\, option\, is\, D. \end{array}$$
  • Question 6
    1 / -0
    The line $$3x-4y=0$$
    Solution
    $$3x=4y$$
    $$x=\dfrac { 4y }{ 3 } $$
    $${ x }^{ 2 }+{ y }^{ 2 }=25$$
    $${ x }^{ 2 }-\left( 1+\dfrac { 16 }{ 9 }  \right) =25$$
    $${ x }^{ 2 }=9$$
    $$x=\pm 3$$
    $$y=\pm 9/4$$
    Now,  $${ x }^{ 2 }+{ y }^{ 2 }=25$$
              $$2x+2y\dfrac { dy }{ dx } =0$$
    $$\dfrac { dy }{ dx } =\dfrac { -x }{ y } =\dfrac { -\left( 3 \right)  }{ \left( 9/4 \right)  } =-4/3$$
    $$\dfrac { -dx }{ dy } =3/4\Rightarrow $$ slope of normal at circle and $$3/4\Rightarrow $$ solpe of $$3x-4y=0$$
    So, Line is normal at circle.
  • Question 7
    1 / -0
    The equation of the normal to the curve $$y^4=ax^3$$ at (a , a) is 
    Solution
    The equation of the curve is,
    $${ y }^{ 4 }=a{ x }^{ 3 }$$

    Differentiate w.r.t x, we get,

    $$4{ y }^{ 3 }\dfrac { dy }{ dx } =3a{ x }^{ 2 }$$

    $$\therefore \dfrac { dy }{ dx } =\dfrac { 3a{ x }^{ 2 } }{ 4{ y }^{ 3 } } $$

    $$\therefore { \dfrac { dy }{ dx }  }_{ \left( a,a \right)  }=\dfrac { 3a\left( a \right) ^{ 2 } }{ 4\left( a \right) ^{ 3 } } $$

    $$\therefore { \dfrac { dy }{ dx }  }_{ \left( a,a \right)  }=\dfrac { 3\left( a \right) ^{ 3 } }{ 4\left( a \right) ^{ 3 } } $$

    $$\therefore { \dfrac { dy }{ dx }  }_{ \left( a,a \right)  }=\dfrac { 3 }{ 4 } $$

    Thus, slope of tangent is $$\dfrac { 3 }{ 4 } $$

    Let $${ m }_{ 1 }$$ = slope of normal

    As tangent and normal are perpendicular to each other, we can write,
    $$\dfrac { 3 }{ 4 } \times { m }_{ 1 }=-1$$

    $$\therefore { m }_{ 1 }=\dfrac { -4 }{ 3 } $$

    Thus, equation of normal passing through $$\left( a,a \right) $$ is,
    $$y-{ y }_{ 1 }={ m }_{ 1 }\left( x-{ x }_{ 1 } \right) $$

    $$\therefore y-a=\dfrac { -4 }{ 3 } \left( x-a \right) $$

    $$\therefore 3\left( y-a \right) =-4\left( x-a \right) $$

    $$\therefore 3y-3a=-4x+4a$$

    $$\therefore 4x+3y=7a$$
  • Question 8
    1 / -0
    If $$x-2y+k=0$$ is a common tangent to $$\displaystyle{ y }^{ 2 }=4x\quad \& \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { 3 } } =1\left( a>\sqrt { 3 }  \right)  $$, then the value of a, k and other common tangent are given by
    Solution
    Given:$$x-2y+k=0$$
    $$\Rightarrow\,2y=x+k$$
    $$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{k}{2}$$
    Comparing with $$y=mx+c$$ we get $$m=\dfrac{1}{2}$$ and $$c=\dfrac{k}{2}$$
    $$\therefore\,y=mx+c$$ is tangent to $${y}^{2}=4ax$$ where $$c=\dfrac{a}{m}$$
    $$\Rightarrow\,{y}^{2}=4x,\,a=1$$
    $$\Rightarrow\,c=\dfrac{1}{m}$$
    $$\Rightarrow\,\dfrac{k}{2}=\dfrac{1}{\dfrac{1}{2}}$$
    $$\Rightarrow\,\dfrac{k}{2}=2$$
    $$\therefore\,k=4$$
    Substituting $$m=\dfrac{1}{2}$$ and $$c=\dfrac{1}{m}=\dfrac{1}{\dfrac{1}{2}}=2$$ in $$y=mx+c$$ we get
    $$\Rightarrow\,y=\dfrac{x}{2}+\dfrac{4}{2}$$
    $$\Rightarrow\,y=\dfrac{x}{2}+2$$
    Given:$$y=mx+c$$ is tangent to $$\dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$$ 
    $$\therefore\,$$ Required condition is $${c}^{2}={a}^{2}{m}^{2}+{b}^{2}$$
    $$\Rightarrow\,4={a}^{2}{\left(\dfrac{1}{2}\right)}^{2}+3$$
    $$\Rightarrow\,\dfrac{{a}^{2}}{4}=4-3=1$$
    $$\Rightarrow\,{a}^{2}=4$$
    $$\Rightarrow\,a=2$$ since $$a>0$$
  • Question 9
    1 / -0
    The normal to the curve, $${{\text{x}}^{\text{2}}}{\text{ + 2xy - 3}}{{\text{y}}^{\text{2}}}{\text{ = 0,}}\;{\text{at}}\;\left( {{\text{1,1}}} \right){\text{:}}$$
    Solution
    Given equation of curve is $${x}^{2}+2xy-3{y}^{2}=0$$ .......$$(1)$$

    On differentiating w.r.t $$x$$ we get

    $$2x+2x{y}^{\prime}+2y-6y{y}^{\prime}=0$$

    $$\Rightarrow\,{y}^{\prime}=\dfrac{x+y}{3y-x}$$

    At $$x=1,\,y=1,\,{y}^{\prime}=1$$

    $$\left(\dfrac{dy}{dx}\right)_\left(1,1\right)=1$$

    Equation of the normal at $$\left(1,1\right)$$ is

    $$y-1=-\dfrac{1}{1}\left(x-1\right)$$

    $$\Rightarrow\,y-1=-x+1$$

    $$\Rightarrow\,x+y=2$$      
     
    $$\Rightarrow\,y=2-x$$     ......$$(2)$$

    On solving equations $$(1)$$ and $$(2)$$,we get

    $${x}^{2}+2x\left(2-x\right)-3{\left(2-x\right)}^{2}=0$$

    $$\Rightarrow\,{x}^{2}+4x-2{x}^{2}-3\left(4+{x}^{2}-4x\right)=0$$

    $$\Rightarrow\,{x}^{2}+4x-2{x}^{2}-12-3{x}^{2}+12x=0$$

    $$\Rightarrow\,-4{x}^{2}+16x-12=0$$

    $$\Rightarrow\,{x}^{2}-4x+3=0$$

    $$\Rightarrow\,\left(x-1\right)\left(x-3\right)=0$$

    $$\Rightarrow\,x=1,\,x=3$$

    When $$x=1,$$then $$y=1$$ using $$(2)$$

    When $$x=3$$ then $$y=-1$$

    $$\therefore\,P\left(1,1\right)$$ and $$Q\left(3,-1\right)$$

    Hence, normal meets the curve again at $$\left(3, - 1\right)$$ in fourth quadrant.
  • Question 10
    1 / -0
    The equation of one of the tangents to the curve $$y=\cos(x+y),-2\pi 
    \le x \le 2\pi$$; that is parallel to the line $$x+ 2y = 0$$ , is
    Solution
    Given curve is
    $$y = \cos (x + y), -2\pi \leq x \leq 2\pi ..... (1)$$
    Parallel to the line
    $$x + 2y = 0$$
    We prove that, slope of tangent is $$\dfrac {dy}{dx}$$
    $$y = \cos (x + y)$$
    Diff w.r.t $$x$$ and we get
    $$\dfrac {dy}{dx} = -\sin (x + y) \dfrac {d}{dx}(x + y) \therefore \dfrac {d}{dx}\cos x = -\sin x$$
    $$\dfrac {dy}{dx} = -\sin (x + y) \left (1 + \dfrac {dy}{dx}\right )$$
    $$\dfrac {dy}{dx} = -\sin (x + y) + (-\sin (x + y))\dfrac {dy}{dx}$$
    $$\dfrac {dy}{dx} [1 + \sin (x + y)] = -\sin (x + y)$$
    $$\dfrac {dy}{x} = \dfrac {-\sin (x + y)}{1 + \sin (x + y)} ..... (2)$$
    Given line is $$x + 2y = 0$$
    $$x = -2y$$
    $$2y = -x$$
    $$y = -\dfrac {x}{2}$$
    $$y = \dfrac {-x}{2} + C$$
    The above equation of the form
    $$y = mx + C$$
    When, $$m = \dfrac {-1}{2}$$ (slope)$$
    $$\dfrac {dy}{dx} = m = \dfrac {-1}{2}$$
    $$\therefore$$ slope of tangent $$=$$ slope of line
    $$\dfrac {-\sin (x + y)}{1 + \sin (x + y)} = -\dfrac {1}{2}$$
    $$\dfrac {\sin (x + y)}{1 + \sin (x + y)} = \dfrac {1}{2}$$
    $$2\sin (x + y) = 1 + \sin (x + y)$$
    $$2 \sin (x + y) - \sin (x + y) = 1$$
    $$\sin (x + y) = 1$$
    $$\sin (x + y) = \sin \dfrac {\pi}{2}$$
    Hence, general solution of $$x + y$$ is
    $$x + y = n\pi + (-1)^{n} \dfrac {\pi}{2}$$
    Now, Finding points through which tangents passes.
    $$y = \cos (x + y)\rightarrow$$ given curve
    Putting value of $$(x + y)$$
    $$y = \cos \left [n\pi + (1)^{n}\dfrac {\pi}{2}\right ]$$
    $$y = 0$$ in $$x$$ (put)
    $$x + y = n\pi + (1)^{n}\dfrac {\pi}{2}$$
    $$x + 0 = n\pi + (-1)^{n} \dfrac {\pi}{2}$$
    $$x = n\pi + (-1)^{n} \dfrac {\pi}{2}$$
    Since, $$-2\pi \leq x \leq 2\pi$$
    Putting $$n = 0$$
    $$x = 0\pi + (-1)^{0} \dfrac {\pi}{2}$$
    $$\therefore (-1)^{0} = 1$$
    $$x = \dfrac {\pi}{2}$$
    Putting $$n = -1$$
    $$x= (-1)\pi + (-1)^{-1} \dfrac {\pi}{2}$$
    $$x = -\pi - \dfrac {\pi}{2}$$
    $$x = -\dfrac {3\pi}{2}$$
    Thus, points are $$\left (-\dfrac {3\pi}{2}, 0\right )$$ and $$\left (\dfrac {\pi}{2}, 0\right )$$
    We know that equation of tangent at point $$(x_{1}, y_{1})$$ and slope $$m$$ is
    $$y - y_{1} = m(x - x_{1})$$
    Now, Equation of tangent at $$\left (\dfrac {-3\pi}{2}, 0\right )$$ and slope $$\dfrac {-1}{2}$$ is
    $$y - 0 = -\dfrac {1}{2} \left (x - \left (\dfrac {-3\pi}{2}\right )\right )$$
    $$y = \dfrac {-1}{2} \left (x + \dfrac {3\pi}{2}\right )$$
    $$4y = -(2x + 3\pi)$$
    $$2x + 4y + 3\pi = 0$$
    Equation of tangent at $$\left (\dfrac {\pi}{2}, 0\right )$$ and slope $$\dfrac {-1}{2}$$ is
    $$y - 0 = \dfrac {-1}{2} \left (x - \dfrac {\pi}{2}\right )$$
    $$y = \dfrac {-1}{2} \left (\dfrac {2x - \pi}{2}\right )$$
    $$4y = -2x + \pi$$
    $$2x + 4y - \pi = 0$$
    Hence, this is the answer.
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