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Tangents and its Equations Test 46

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Tangents and its Equations Test 46
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  • Question 1
    1 / -0
    The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is :
    Solution

    $${\textbf{Step -1: Differentiate slope}}{\textbf{.}}$$

                     $${\text{Given: slope of tangent is equal to ratio of abscissa to the ordinate of the point}}{\text{.}}$$

                     $${\text{Let slope of the tangent be }}\dfrac{{dy}}{{dx}}.$$

                     $$\text{According to the question}$$

                     $$\dfrac{{dy}}{{dx}} = \dfrac{x}{y}$$

    $${\textbf{Step -2: Solve further to find the equation}}{\textbf{.}}$$

                     $${\text{Separating the variables}}{\text{.}}$$

                     $$ydy = xdx$$

                     $${\text{Upon integration we get,}}$$

                     $$\int {ydy = \int {xdx} } $$

                     $$\Rightarrow\dfrac{{{y^2}}}{2} = \dfrac{{{x^2}}}{2} + C$$

                     $$\Rightarrow{y^2} = {x^2} + 2C$$

                     $$ \Rightarrow {x^2} - {y^2} + 2C = 0$$

                     $$ \Rightarrow {x^2} - {y^2} = k$$            $$\textbf{[  k = - 2 C]} $$

    $${\textbf{Hence, the given equation is of a rectangular hyperbola}}{\textbf{.}}$$

  • Question 2
    1 / -0
    The tangent to the curve $$ y=e^{2x} $$ at the point $$ (0,1) $$ meets x-axis at:
    Solution
    The equation of curve is $$ y=e^{2x} $$
    Since, it passes through the point $$ (0.1).$$
    $$ \therefore \frac{dy}{dx}=e^{2x}.2=2.e^{2x} $$
    $$ \Rightarrow \left( \frac { dy }{ dx }  \right) _{ x=0 }=2e^{2\times0} =2= $$ slope of tangent to the curve
    $$ \therefore $$ Equation of tangent is $$ y-1=2(x-0) $$
    $$ \Rightarrow y=2x+1 $$
    Since, tangent to curv $$ y=e^{2x} $$ at the point $$ (0,1) $$ meets X-axis i.e., $$ y=0 $$.
    $$ \therefore 0=2x+1 \Rightarrow x= -\frac{1}{2} $$
    So, the required point is $$ \left( \frac { -1 }{ 2 } ,0 \right) . $$
  • Question 3
    1 / -0
    The equation of tangents to the curve $$ y(1+x^2 )=2-x, $$ where it crosses x-axis is:
    Solution
    We have, equation of the curve $$ y(1+x^2)=2-x...(i) $$
    $$ \therefore y.(0+2x)+(1+x^2).\frac{dy}{dx}=0-1 $$ [on differentiating w.r.t.x]
    $$ \Rightarrow 2xy+ (1+x^2)\frac{dy}{dx}=-1$$
    $$ \Rightarrow \dfrac{dy}{dx} = \dfrac{-1-2xy}{1+x^2}...(ii) $$
    Since, the given curve passes through $$ x-$$ axis i.e., $$ y=0.$$
    $$ \therefore 0(1+x^2)=2-x$$ [using Eq.(i)]
    $$ \Rightarrow x=2 $$
    So, the curve passes through the point $$ (2,0). $$ 
    $$ \therefore \left( \frac { dy }{ dx }  \right) _{ (2,0) }=\dfrac { -1-2\times 0 }{ 1+2^{ 2 } } =-\dfrac { 1 }{ 5 } = $$ slope of the curve
    $$ \therefore $$ slope of tangent to the curve = $$ - \frac{1}{5} $$
    $$ \therefore $$ Equation of tangent of the curve passing through $$ (2,0) $$ is
    $$ y-0=-\frac{1}{5}(x-2) $$
    $$ \Rightarrow 5y=-x+2 $$
    $$ \Rightarrow 5y+x=2 $$
  • Question 4
    1 / -0
    The slope of tangent to the curve $$ x=t^2+3t-8,y=2t^2-2t-5 $$ at the point $$ (2,-1) $$ is:
    Solution
    Equation of curve is given is given by
    $$ x=t^2+3t-8 $$ and $$ y=2t^2-2t-5. $$
    $$ \therefore \dfrac{dx}{dt}= 2t+3 $$ and $$ \dfrac{dy}{dx}=4t-2 $$
    $$ \Rightarrow\dfrac{dy}{dx}=\dfrac { \frac { dy }{ dt }  }{ \frac { dx }{ dt }  } =\dfrac { 4t-2 }{ 2t+3 } ....(i) $$
    Since, the curve passes through the point $$ (2,-1). $$
    $$ \therefore 2=t^2+3t-8 $$
    and $$ -1=2t^2-2t-5 $$
    $$ \Rightarrow t^2+3t-10=0 $$
    and $$ 2t^2-2t-4=0 $$
    $$ \Rightarrow t^2+5t-2t-10=0 $$  
    and $$ 2t^2+2t-4t-4=0 $$
    $$ \Rightarrow t(t+5)-2(t+5)=0 $$
    and $$ 2t(t+1)-4(t+1)=0 $$
    $$\Rightarrow (t-2)(t+5)=0 $$
    and $$ (2t-4)(t+1)=0 $$
    $$ \Rightarrow t=2,-5 $$ and $$ t=-1,2 $$
    $$ \Rightarrow t=2 $$
    $$ \therefore $$ Slope of tangent,
    $$ \left( \frac { dy }{ dx }  \right) _{ at\ t-2 }=\dfrac { 4\times 2-2 }{ 2\times 2+3 } =\dfrac { 6 }{ 7 } $$ [using Eq.(i)] 
  • Question 5
    1 / -0
    The equation of the tangent to the curve $$y=1-e^{\dfrac{x}{2}}$$ at the point of intersection with $$Y-$$ axis 
  • Question 6
    1 / -0
    The line $$5x-2y+4k=0$$ is tangent to $$4x^{2}-y^{2}=36$$, then k is:
    Solution
    The equation of the hyperbola is
    $$4x^2-y^2=36$$
    $$\dfrac{4x^2}{36}-\dfrac{y^2}{36}=1$$
    $$\dfrac{4x^2}{9}-\dfrac{y^2}{36}=1$$............(1)
    $$a^2=9;$$
    $$b^2=36$$
    The equation of the line is
    $$5x-2y+4k=0$$
    $$2y=5x+4k$$
    $$y=\dfrac{5}{2}x+2k$$............(2)
    From equation (2) $$m=\dfrac{5}{2},\,c=2k$$
    $$c^2=a^2m^2-b^2$$
    $$(2k)^2=9(\dfrac{5}{2})^2-36$$
    $$4k^2=9\times \dfrac{25}{4}-36$$
    $$4k^2=\dfrac{225-144}{4}$$
    $$k^2=\dfrac{81}{16}\Rightarrow k=\dfrac{9}{4}$$
  • Question 7
    1 / -0
    If the tangent at $$(1,1)$$ on $$y^{2}=x(2-x)^{2}$$ meets the curve again at $$P$$, then $$P$$ is
    Solution
    $$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$$
    [Hint: $$y^{2}=x(2-x)^{2}=x(4-4x+x^{2})=x^{3}-4x^{2}-4x$$
    $$\therefore 2y\dfrac{dy}{dx}=3x^{2}-8x+4$$
    $$\therefore \dfrac{dy}{dx}=\dfrac{3x^{2}-8x+4}{2y}$$
    $$\therefore \left(\dfrac{dy}{dx}\right)_{at\ (1,1)}=\dfrac{3(1)^{2}-8(1)+4}{2(1)}=\dfrac{1}{2}$$
    = slope of the tangent at $$(1,1)$$
    $$y-1=-\dfrac{1}{2}(x-1)$$
    $$\therefore 2y-12=-x+1$$
    $$\therefore x+2y=3$$
    Only the coordinates $$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$$ satisfy both the 
    equation $$y^{2}=x(2-x)^{2}$$ and $$x+2y=3$$
    $$\therefore P$$ is $$\left(\dfrac{9}{4}, \dfrac{3}{8}\right)$$ ].
  • Question 8
    1 / -0
    The slope of the tangent to the curve $$x = t^{2} + 3 t - 8, y = 2t^{2} - 2t - 5$$ at the point $$(2, -1)$$ is
    Solution
    $$Slope \space of \space tangent \space to \space curve \space y=f(x) \space is \dfrac{dy}{dx}$$
    $$Using\space chain \space rule$$
    $$\dfrac{dy}{dx} = \dfrac{(\dfrac{dy}{dt})}{(\dfrac{dx}{dt})}$$ 
    $$Given\space x=t^2+3t-8 \implies \dfrac{dx}{dt}=2t+3$$
    $$Given\space y=2t^2-2t-5 \implies \dfrac{dy}{dt}=4t-2$$
    $$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{4t - 2}{2t + 3}$$
    $$\dfrac{dy}{dx} |_{t=2} = \dfrac{4 \times 2 - 2}{2 \times 2 + 3} = \dfrac{6}{7}$$
    $$\therefore $$Answer (B) $$\dfrac{6}{7}$$
  • Question 9
    1 / -0
    The normal at the point $$(1, 1)$$ on the curve $$2y + x^{2} - 3$$ is .............
    Solution
    Given, $$2y + x^{2} = 3$$
    $$2 \dfrac{dy}{dx} + 2x = 0 \\ \Rightarrow \dfrac{dy}{dx} = -x \\ \therefore \dfrac{dy}{dx} = -1$$
    Slope of tangent = $$-1$$
    $$\therefore $$ equation of normal is $$y - 1 = 1 (x - 1)$$
    $$\Rightarrow y - x = 0$$
    $$\Rightarrow  x-y = 0$$
    $$\therefore $$ Answer (B).)
  • Question 10
    1 / -0
    The normal to the curve $$x^{2} = 4y$$ passing $$(1, 2)$$ is
    Solution
    Given, $$4y = x^{2} \Rightarrow 4 \dfrac{dy}{dx} = 2x \Rightarrow \dfrac{dy}{dx}|_{x = x1} = \dfrac{x_{1}}{2}$$
    slope of tangent is $$ = \dfrac{x_{1}}{2}$$
    slope of normal is $$= \dfrac{-2}{x_{1}}$$
    equation of normal $$y - y_{1} = \dfrac{-2}{x_{1}} (x - x_{1})$$ 
    but this passes they $$(1, 2)$$
    $$2 - y_{1} = \dfrac{-2}{x} (1 - x) \Rightarrow 2 - y_{1} = \dfrac{-2}{x_{1}} + 2$$
    $$y_{1} = \dfrac{2}{x_{1}}$$
    but $$x^{2} = 4y$$
    $$x^{2}_{1} = 4y_{1}$$
    $$x^{2}_{1} = 4 \times \dfrac{2}{x_{1}} = 8  \Rightarrow x_{1} = 2$$
    when $$x_{1} = 2 y_{1} = \dfrac{2}{2} = 1$$
    $$\therefore $$ equation of normal is
    $$y - 1 = \dfrac{-2}{2} \times (x - 2) \Rightarrow y - 1 = (x - 2)$$
    $$x + y = 3$$
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