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Tangents and its Equations Test 49

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Tangents and its Equations Test 49
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  • Question 1
    1 / -0
    If the circle $$x^2+y^2+2gx+2fy+c=0$$ is touched by $$y=x$$ at P such that OP = $$6\sqrt{2}$$
    then the value of c is
    Solution
    Let the point of contact be $$x_{1},y_{1}$$
    However it lies on the line $$y=x$$
    Hence
    $$x_{1}=y_{1}$$
    Applying distance formula, we get
    $$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
    $$2x_{1}^2=72$$
    $$x_{1}=6$$ ...(positive, since it lies on y=x.)
    Differentiating the equation of circle with respect to x.
    $$2x+2yy'+2g+2fy'=0$$
    Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
    By substituting, we get
    $$x+y=-(g+f)$$
    Now $$x_{1}=y_{1}=6$$
    Hence
    $$12=-(g+f)$$ ...(i)
    Substituting $$x=y=6$$ in the equation of the circle, we get
    $$36+36+2(6)(g+f)+c=0$$
    $$72+12(-12)+c=0$$
    $$c=144-72$$
    $$c=72$$
  • Question 2
    1 / -0
    The angle made by the tangent of the curve $$\displaystyle x = a(t + \sin t \cos t); y = a (1 + \sin t)^2$$ with the x-axis at any point on it is
    Solution
    $$x=a\left(t+\sin{t}\cos{t}\right)$$
    $$\dfrac{dx}{dt}=a\left(1+{\cos}^{2}{t}-{\sin}^{2}{t}\right)$$
    $$\quad \ =a\left(1+\left({\cos}^{2}{t}-{\sin}^{2}{t}\right)\right)$$
    $$\dfrac{dx}{dt}=a\left(1+\cos{2t}\right)$$

    $$y=a{\left(1+\sin{t}\right)}^{2}$$
    $$\dfrac{dy}{dt}=2a\left(1+\sin{t}\right)\cos{t}$$
    $$\dfrac{dy}{dt}=2a\cos{t}\left(1+\sin{t}\right)$$

    $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\$$
    $$=\dfrac{2a\cos{t}\left(1+\sin{t}\right)}{a\left(1+\cos{2t}\right)}\\$$
    $$=\dfrac{2\cos{t}\left(1+\sin{t}\right)}{2{\cos}^{2}{t}}\\$$
    $$=\dfrac{\left(1+\sin{t}\right)}{\cos{t}}\\$$
    $$=\dfrac{\left({\cos}^{2}{\dfrac{t}{2}}+{\sin}^{2}{\dfrac{t}{2}}+2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}\right)}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$$
    $$=\dfrac{{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}^{2}}{{\cos}^{2}{\dfrac{t}{2}}-{\sin}^{2}{\dfrac{t}{2}}}\\$$
    $$=\dfrac{\left(\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}\right)}{\left(\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}\right)}\\$$
    $$=\dfrac{1+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{t}{2}}}\\$$
    $$=\dfrac{\tan{\dfrac{\pi}{4}}+\tan{\dfrac{t}{2}}}{1-\tan{\dfrac{\pi}{4}}\tan{\dfrac{t}{2}}}\\$$
    $$=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}\\$$

    If $$\theta$$ is the angle which the tangent at any point $$t$$ makes with the $$X$$ axis
    then $$\tan{\theta}=\tan{\left(\dfrac{\pi}{4}+\dfrac{t}{2}\right)}$$

    $$\therefore\,\theta=\dfrac{\pi}{4}+\dfrac{t}{2}=\dfrac{1}{4}\left(\pi+2t\right)$$
  • Question 3
    1 / -0
    Equation of the line through the point $$\left(\dfrac{1}{2}, 2 \right)$$ and tangent to the parabola $$\displaystyle y = \frac {-x^2}{2}+2$$ and secant to the curve $$\displaystyle y = \sqrt {4 - x^2}$$ is :
    Solution
    We have, 
    $$y = \dfrac{-x^2}{2} +2 $$
    $$y' = -x$$
    At $$(x_1,y_1)$$, 
    $$y' = -x_1$$
    The equation of a tangent at $$(x_1, y_1)$$ can be written as $$ y - y_1 = -x_1 (x- x_1)$$
    The tangent passes through $$\left(\dfrac12,2 \right)$$.
    Hence, 
    $$ y_1 -2 = -x_1 \left(x_1 - \dfrac12 \right) $$
    $$ -\dfrac{{x_1}^2}{2} = -{x_1}^2 + \dfrac{x_1}{2} $$
    $$ x_1 = x_1^2$$
    $$x_1 = 0,1$$
    The equation at $$x_1= 1, y_1 = \dfrac32$$, 
    $$y -\dfrac32 = -1(x-1) $$
    $$ 2x+2y -5 = 0$$. The distance of the line from the origin is $$ \dfrac{5}{2\sqrt{2}} <2 $$. Hence, this line is a secant to the circle. 
    The tangent at $$(2,0)$$ is given by $$x=2$$. However this is also tangent to the circle. 
    Hence, option A is correct.
  • Question 4
    1 / -0
    What is the minimum intercept made by the axes on the tangent to the ellipse $$ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ ?
    Solution
    Equation of ellipse is$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
    Let $$P (a\cos\theta, b\sin\theta)$$ be a point on the ellipse.Equation of tangent to ellipse at P is $$\dfrac{x}{a}{\cos\theta}+\dfrac{y}{b}{\sin\theta}=1$$
    So, the tangent intersects X-axis at $$(a\sec\theta,0)$$ and Y-axis at $$(0,bcosec\theta)$$
    Intercepted portion of tangent between axes is $$S=\sqrt{a^2\sec^{2}\theta+b^2cosec^{2}\theta}$$
    For maximum or minimum , $$\dfrac{dS}{d\theta}=0$$$$\Rightarrow 2a^2\sec^{2}\theta\tan\theta-2b^2cosec^{2}\theta\cot\theta=0$$$$\Rightarrow \tan^{4}\theta=\dfrac{b^2}{a^2}$$$$\Rightarrow \tan^{2}\theta=\dfrac{b}{a}$$
    Also, $$\dfrac{d^2S}{d\theta^2}>0$$ for $$\tan^{2}\theta=\dfrac{b}{a}$$
    Hence, S attains minimum at $$\tan^{2}\theta=\dfrac{b}{a}$$
    Minimum value of $$S=\sqrt{a^2(1+\tan^{2}\theta)+b^2(1+\cot^{2}\theta)}$$$$\Rightarrow S_{min}=\sqrt{a(a+b)+b(a+b)}$$
    Minimum value of S $$=a+b$$
  • Question 5
    1 / -0
    Let $$C$$ be the curve $$\displaystyle y = x^3$$ (where $$x$$ takes all real values.) The tangent at $$A$$ meets the curve again at $$B$$. If the gradient of the curve at $$B$$ is $$K$$ times the gradient at $$A$$ then $$K$$ is equal to
    Solution
    Given equation of curve is $$y=x^{3}$$
    Let the coordinates of A be $$(t,t^{3})$$ and coordinates of B be $$(T,T^{3})$$
    $$\displaystyle \dfrac{dy}{dx}=3x^{2}$$
    Slope of tangent to curve at A $$= 3t^{2}$$.

    Now , slope of AB $$=\displaystyle \dfrac{T^3-t^3}{T-t}$$

    $$\Rightarrow \displaystyle \dfrac{T^3-t^3}{T-t} =3t^{2}$$

    $$\Rightarrow {T^3-t^3}=3t^{2}(T-t)$$

    $$\Rightarrow (T-t)(T^2+tT+t^2)-3t^{2}(T-t)=0$$

    $$\Rightarrow (T-t)(T^2+tT-2t^2)=0$$

    $$\Rightarrow T=t, T=-2t$$

    $$T=t$$ is not possible. Hence, $$T=-2t$$

    Given, $$ m_B =K\cdot m_A$$
    $$\Rightarrow T^2=Kt^{2}$$
    $$\Rightarrow 4t^{2}=Kt^{2}$$
    $$\Rightarrow K=4$$
  • Question 6
    1 / -0
    The graphs $$y=2x^3-4x+2$$ and $$y=x^3+2x-1$$ intersect at exactly 3 distinct points. The slope of the line passing through two of these points
    Solution
    Given equation of curves 
    $$y=2x^3-4x+2$$
    $$y=x^3+2x-1$$ 
    Let $$(x_{1},y_{1})$$ be a point of intersection of the curves.
    $$\Rightarrow y_{1}=2x_{1}^3-4x_{1}+2$$ 
    And $$ y_{1}=x_{1}^3+2x_{1}-1$$ 
    $$\Rightarrow y_1=8x_{1}-4$$
    Let $$(x_{2},y_{2})$$ be another point of intersection of the curves.
    $$\Rightarrow y_{2}=2x_{2}^3-4x_{2}+2$$ 
    And $$y_{2}=x_{2}^3+2x_{2}-1$$ 
    $$\Rightarrow y_2=8x_{2}-4$$
    Now, slope $$= \displaystyle \dfrac{y_2-y_1}{x_2-x_1}$$
    $$\Rightarrow m =\displaystyle \dfrac{8(x_2-x_1)}{x_2-x_1}$$
    $$\Rightarrow m=8$$
  • Question 7
    1 / -0
    Consider the curve represented parametrically by the equation
    $$\displaystyle x = t^3 - 4t^2 - 3t$$ and $$\displaystyle y = 2t^2 + 3t - 5$$ where $$\displaystyle t \: \epsilon \: R$$.
    If $$H$$ denotes the number of point on the curve where the tangent is horizontal and $$V$$ the number of point where the tangent is vertical then
    Solution
    $$\dfrac{dx}{dt} = 3t^2-8t-3$$
    $$\dfrac{dy}{dt} = 4t+3$$
    $$\Rightarrow \dfrac{dy}{dt} = \dfrac{4t+3}{3t^2-8t-3}$$
    clerarly denominator has two roots and numerator has single root
    Hence given curve will have two vertical tangent and one horizontal tangent
    $$\Rightarrow H =1 , V =2$$ 
  • Question 8
    1 / -0
    The number of points on the curve $$x^{3/2}+y^{3/2}=a^{3/2}$$, where the tangents are equally inclined to the axes, is
    Solution
    If the tangents are equally inclined to the axes we get $$y'_{x,y}=1$$.
    Therefore 
    $$\dfrac{3}{2}\sqrt{x}+\dfrac{3}{2}\sqrt{y}.y'=0$$

    $$y'=1$$ implies 
    $$\sqrt{x}+\sqrt{y}=0$$

    $$\sqrt{x}=-\sqrt{y}$$

    $$x=y$$.
    $$x_{1}=y_{1}$$.
    Substituting in the equation we get 
    $$2x^{\frac{3}{2}}=a^{\frac{3}{2}}$$

    $$x=\dfrac{a}{2^{\frac{2}{3}}}$$
    Therefore 
    $$(x_{1},y_{1})=\left(\dfrac{a}{2^{\frac{2}{3}}},\dfrac{a}{2^{\frac{2}{3}}}\right)$$
    Therefore there exists only one point.
  • Question 9
    1 / -0
    Let $$\displaystyle f(x) = ln \: mx (m > 0)$$ and $$g(x) = px$$. Then the equation $$\displaystyle |f(x)| = g(x)$$ has only one solution for
    Solution
    Let $$P(h,k)$$ be a point on $$f(x)=\ln mx$$
    then, $$k=\ln mh$$
    $$\dfrac { dy }{ dx } $$ at point P $$=\dfrac { 1 }{ h } $$
    Equation of tangent at point P $$L_{1}: y-k=\dfrac { \left( x-h \right)  }{ h } $$
    Since, it passes through origin
    Therefore, $$k=1$$ & $$h=e/m$$
    and $$L_{1}: y=x/h$$    $$\Rightarrow L_{1}: y=mx/e$$

    From the figure: $$|f(x)|=g(x)$$ have one solution when slope of $$g(x)$$ is greater than slope of $$L_{1}$$
    Therefore, $$p>m/e$$

    Ans: D

  • Question 10
    1 / -0
    A point $$\displaystyle P(a, a^n)$$ on the graph of $$\displaystyle y = x^n (n \: \epsilon \: N)$$ in the first quadrant a normal is drawn. The normal intersects the y-axis at the point $$(0, b)$$. If $$\displaystyle _{a \rightarrow 0}^{Lim \: b} \textrm{=} \frac {1}{2}$$, then n equals
    Solution
    $$y={ x }^{ n }$$
    Slope of normal is $$-\cfrac { dx }{ dy } =-\cfrac { 1 }{ n{ \alpha  }^{ n-1 } } $$
    So the equation of normal is 
    $$\left( y-{ \alpha  }^{ n } \right) =-\cfrac { 1 }{ n{ \alpha  }^{ n-1 } } \left( x-\alpha  \right) \\ y+\cfrac { x }{ n{ \alpha  }^{ n-1 } } =n{ \alpha  }^{ 2-n }+{ \alpha  }^{ n }$$
    Therefore $$b=n{ \alpha  }^{ 2-n }+{ \alpha  }^{ n }$$
    Hence to exists $$\displaystyle\lim _{ \alpha \rightarrow 0 }{ b } $$ must be equal to $$2$$
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