Tangents and its Equations Test 48

At any point $$(k, k^2+8)$$ on $$y=x^2+8$$, the slope of tangent is $$ \dfrac{dy}{dx}|_{x=k}=2k$$.....(1)

Similarly, at any point $$(p, -p^2)$$ on $$y=-x^2$$, the slope of tangent is $$ \dfrac{dy}{dx}|_{x=p}=-2p$$....(2)

For common tangent, $$2k=-2p \Rightarrow k=-p$$

Hence, the coordinates of point on common tangent and on $$y=-x^2$$ can be written as $$(-k, -k^2)$$ as shown in figure.

So, slope of common tangent is $$ \dfrac{8+k^2-(-k^2)}{k-(-k)}=\dfrac{8+2k^2}{2k}$$

From (1), slope is also equal to $$2k$$.

Hence, $$\dfrac{8+2k^2}{2k}=2k$$

Solving, we get $$k= \pm 2$$.

Discarding negative value (since 'k' is positive as defined in the function), we get $$k=2$$

Hence, slope of common tangent is $$2k=4$$

The point common to tangent and $$y=x^2+8$$ is $$(2,12)$$

Hence, equation of common tangent is $$y-12=4(x-2)$$

To find $$x$$ intercept, substitute $$y=0$$ in the above equation.

We get $$x$$ intercept$$=-1$$

$$f''(x)=6x-6$$
$$\int { f''(x)dx } =\int { (6x-6)dx } $$
$$\Rightarrow f'(x)=3x^{2}-6x+C$$
Since, it passes through (2,1) and tangent is $$y=3x-5$$
$$\Rightarrow C=3$$
So, $$f'(x)=3x^{2}-6x+3$$
$$\int { f'(x)dx } =\int { (3{ x }^{ 2 }-6x+3)dx } $$
$$\Rightarrow f(x)=x^{3}-3x^{2}+3x+C_1$$
Since, it passes through $$ (2,1)$$
$$\Rightarrow C_1=-1$$
Hence, $$f(x)=x^{3}-3x^{2}+3x-1$$
$$f(0)=-1$$

$$y=e^{x}+e^{-x}$$
$$\displaystyle\frac{dy}{dx}=e^{x}-e^{-x}=\frac{e^{2x}-1}{e^{x}}$$
Let P$$(x_{1},y_{1})$$ and Q$$(x_2,y_2)$$ be the points on the given curve.
Slope of tangent at P is $$\displaystyle m_1=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$$
$$\Rightarrow \displaystyle \sqrt{3}=\frac{e^{2x_{1}}-1}{e^{x_{1}}}$$
$$\displaystyle\Rightarrow e^{2x_1}-\sqrt{3}e^{x}-1=0$$
$$\displaystyle\Rightarrow (e^{x_1}-\frac{\sqrt{3}}{2})^{2}=\frac{7}{4}$$
$$\displaystyle\Rightarrow e^{x_1}=\pm\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}$$
$$\displaystyle\Rightarrow {x_1}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$$ as logarithm of negative numbers is not defined.
Similarly, $$\displaystyle\Rightarrow {x_2}=ln{(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2})}$$ as it makes same angle,$$60^{0}$$ with x-axis.

Given equation of curve is $$y^{2} = x$$
$$\displaystyle \frac{dy}{dx}=\frac{1}{2y}$$
Slope of tangent $$=\dfrac{1}{2y}$$
Also, given slope of tangent $$=tan 45^{\circ}$$
$$\displaystyle\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$$
$$\displaystyle \Rightarrow x=\frac{1}{4}$$
Hence, the required point is  $$\left ( \dfrac{1}{4},\dfrac{1}{2} \right )$$

$$x=2 \ln\cot t+1$$
$$\displaystyle \frac{dx}{dt}=-\frac{2cosec^{2}t}{\cot t}=-\frac{2}{\sin t\cos t}$$
 $$y=\tan t+ \cot t$$
$$\displaystyle \frac{dy}{dt}=sec^{2}t-cosec^{2}t=\frac{sin^{2}t-cos^{2}t}{sin^{2}t cos^{2}t}$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=-\frac{sin^{2}t-cos^{2}t}{sin2t}$$
$$\Rightarrow \displaystyle \frac{dy}{dx}=\cot 2t$$
Slope of tangent at $$t=\dfrac{\pi}{4}$$ is 0.
Hence, tangent is parallel to x-axis.

$$x=\sec^{2}t$$
$$y=\cot t$$
$$\dfrac{dy}{dx}=-\dfrac{\csc^{2}t}{2\sec^{2}t\cdot\tan t}=-\dfrac{\cot^{3}t}{2}$$

let $$P \left( \dfrac { \sqrt { a }  }{ \cos { \theta  }  } ,\dfrac { \sqrt { b }  }{ \sin { \theta  }  }  \right) $$ be any point of curve $$\dfrac { a }{ x^{ 2 } } +\dfrac { b }{ y^{ 2 } } =1$$
$$\dfrac { -2a }{ x^{ 3 } } -\dfrac { 2b }{ y^{ 3 } } \dfrac { dy }{ dx } =0$$
$$\dfrac { dy }{ dx } $$at point P$$=-\sqrt { \dfrac { b }{ a }  } \cot ^{ 3 }{ \theta  } $$
Equation of tangent at point P: $$\displaystyle y-\frac { \sqrt { b }  }{ \sin { \theta  }  } =-\sqrt { \frac { b }{ a }  } \cot ^{ 3 }{ \theta  } \left( x-\frac { \sqrt { a }  }{ \cos { \theta  }  }  \right) $$
for X-intercept put $$x=0$$
$$\displaystyle 0-\frac { \sqrt { b }  }{ \sin { \theta  }  } =-\sqrt { \frac { b }{ a }  } \cot ^{ 3 }{ \theta  } \left( x-\frac { \sqrt { a }  }{ \cos { \theta  }  }  \right) $$
$$\displaystyle \Rightarrow x=\sqrt { a } \left( \frac { \sin ^{ 2 }{ \theta  }  }{ \cos ^{ 3 }{ \theta  }  } +\frac { 1 }{ \cos { \theta  }  }  \right) =\frac { \sqrt { a }  }{ \cos ^{ 3 }{ \theta  }  } $$
Therefore, X-intercept is proportional to cube of abcissa.

Ans: C