Tangents and its Equations Test 30

Given, curve is $$y=x^3 - 3x$$     .....(i)
On differentiating w.r.t. $$x$$, we get 
$$ \dfrac {dy}{dx} = 3x^2 - 3 $$
Since, tangent is parallel to $$x$$-axis.
Therefore, $$  \dfrac {dy}{dx} = 0 $$
$$\Rightarrow  3x^2 - 3 = 0$$

We have
$$\Rightarrow { y }^{ 2 }={ x }^{ 3 }-4{ x }^{ 2 }+4x$$    .....(i)

Given, $$y=a{ x }^{ 2 }+bx$$
On differentiating with respect to $$x$$, we get
$$\dfrac { dy }{ dx } =2ax+b$$
At $$\left( 2,-8 \right) ,$$  $${ \left( \dfrac { dy }{ dx }  \right)  }_{ \left( 2,-8 \right)  }=4a+b$$
Since tangent is parallel to $$X$$-axis.
Thus $$ \dfrac { dy }{ dx } =0\Rightarrow b=-4a$$             ....(i)
Now, point $$\left( 2,-8 \right)$$ is on the curve $$ y=a{ x }^{ 2 }+bx$$
Therefore, $$ -8=4a+2b$$            ....(ii)
On solving equations (i) and (ii), we get
$$a=2,b=-8$$

Equation of tangent is $$y-2x+1=0$$

It is tangent at $$x=1$$, so for $$x=1$$

$$y-2(1)+1=0\\ \Rightarrow y=1$$

So, its is tangent to the curve at $$(1,1)$$

Slope of tangent $$= -\left (\dfrac{-2}{1}\right)=2$$

$$xy+ax+by=0\\ y+x\dfrac { dy }{ dx } +a+b\dfrac { dy }{ dx } =0$$

Now $$\dfrac { dy }{ dx } =2$$ at $$(1,1)$$

$$1+1(2)+a+b(2)=0\\ \Rightarrow a+2b=-3$$     .....(i)

$$(1,1)$$ also lies on the curve $$xy+ax+by=0$$

$$\Rightarrow 1(1)+a(1)+b(1)=0\\ \Rightarrow a+b=-1$$     .......(ii)

Solving (i) and (ii), we get

$$\Rightarrow a=1,b=-2$$

So, option E is correct.

Given curves are
$$x = t^{2} + 3t - 8$$
$$\therefore \dfrac {dx}{dt} = 2t + 3$$
and $$y = 2t^{2} - 2t - 5$$
$$\therefore \dfrac {dy}{dt} = 4t - 2$$
Slope of tangent $$= \dfrac {dy}{dx} = \dfrac {dy}{dx} \times \dfrac {dt}{dx} = \dfrac {4t - 2}{2t + 3} .... (i)$$
Since, curve passes through the point $$(2, -1)$$.
Therefore, $$t^{2} + 3t - 8 = 2$$ and $$2t^{2} - 2t - 5 = -1$$
$$\Rightarrow t^{2} + 3t - 10 = 0$$
and $$2t^{2} - 2t - 4 = 0$$
$$\Rightarrow t^{2} + 5t - 2t - 10 = 0$$
and $$t^{2} - t - 2 = 0$$
$$\Rightarrow (t + 5)(t - 2) = 0$$ and $$(t^{2} - 2t + t - 2) = 0$$
$$\Rightarrow t = -5, 2$$ and $$(t - 2)(t + 1) = 0$$
$$\Rightarrow t = -5, 2$$ and $$t = -1, 2$$
So, common value of $$t$$ is $$2$$.
On putting $$t = 2$$ in Eq. (i), we get
$$\left [\dfrac {dy}{dx}\right ]_{at\ t = 2} = \dfrac {4(2) - 2}{2(2) + 3} = \dfrac {6}{7}$$
Slope of normal $$= \dfrac {-1}{\dfrac {dy}{dx}} = \dfrac {-1}{(6/7)} = -\dfrac {7}{6}$$.

Given curves are $$ y = 2^x $$ and $$ y =3^x $$
The point of intersection is $$ 3^x = 2^x \Rightarrow x = 0 $$
On differentiating w.r.t. $$x,$$ we get 
$$ \dfrac {dy}{dx} = 2^x \log 2 = m_1 $$
and $$ \dfrac {dy}{dx} = 3^x \log 3 = m_2 $$
Therefore, $$  \tan \alpha = \dfrac {m_2 - m_1}{1 + m_1m_2} $$
$$ = \dfrac {3^x \log 3 - 2^x \log 2 }{1+3^x \times 2^x \log 3 \times \log 2} $$
At $$x=0$$, 

Given cure is $$ y^3 - xy - 8 = 0 $$
On differentiating w.r.t. $$x,$$ we get 
$$ 3y^2 \dfrac {dy}{dx} - x \dfrac {dy}{dx} - y - 0 =0 $$
$$ \Rightarrow \dfrac {dy}{dx} (3y^2 - x) = y $$
$$ \Rightarrow \dfrac {dy}{dx} = \dfrac {y} {(3y^2 - x) } $$
$$ \Rightarrow \left( \dfrac  {dy}{dx} \right)_{(0,2)} = \dfrac {2}{3(2)^2 - 0} = \dfrac {1}{6} $$
Therefore, the slope of the normal is $$  - \dfrac {1}{dy/dx} = -6 $$.

Given curve is
$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1 $$    ... (i)
On differentiating w.r.t. to $$x$$, we get
$$\dfrac {1}{\sqrt {a}} \cdot \dfrac {1}{2\sqrt {x}} + \dfrac {1}{\sqrt {b}} \cdot \dfrac {1}{2\sqrt {y}} \dfrac {dy}{dx} = 0$$
$$\Rightarrow \dfrac {dy}{dx} = -\dfrac {\sqrt {b}\sqrt {y}}{\sqrt {a}\sqrt {x}} \Rightarrow \left [\dfrac {dy}{dx}\right ]_{(x_{1}, y_{1})} = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}}$$
Equation of tangent passing through the point $$(x_{1}, y_{1})$$ is
$$(y - y_{1}) = \dfrac {-\sqrt {b}\sqrt {y_{1}}}{\sqrt {a}\sqrt {x_{1}}} (x - x_{1})$$
$$\dfrac {y}{\sqrt {by_{1}}} - \dfrac {y_{1}}{\sqrt {by_{1}}} = -\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {x_{1}}{\sqrt {ax_{1}}}$$
$$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = \dfrac {x_{1}}{\sqrt {ax_{1}}} + \dfrac {y_{1}}{\sqrt {by_{1}}} = \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}}$$
$$\Rightarrow \dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = 1$$ [from Eq. (i)]
$$\left [\because at (x_{1}, y_{1}), \sqrt {\dfrac {x_{1}}{a}} + \sqrt {\dfrac {y_{1}}{b}} = 1\right ]$$
But $$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$$ (given)
Therefore, $$k = 1$$

Given curve is $$y = x^2 - \dfrac{1}{x^2}$$
On differentiating both sides w.r.t. $$x$$, we get
$$\dfrac{dy}{dx} = 2x + \dfrac{2}{x^3}$$
At point $$(-1, 0)$$.
$$\dfrac{dy}{dx} = 2(-1) + \dfrac{2}{(-1)^3} = -4$$
Therefore, slope of normal to the curve
$$= - \dfrac{1}{dy/dx}$$
$$= - \dfrac{1}{-4} = \dfrac{1}{4}$$

Given curve is
$$y = 5 + x - x^{2}$$     ..... (i)
On differentiating w.r.t. $$x$$, we get
$$\dfrac {dy}{dx} = 1 - 2x$$
Slope of normal $$= \dfrac {-1}{(dy/dx)} = \dfrac {-1}{1 - 2x}$$
$$= \dfrac {1}{2x - 1}$$     .... (ii)
Since, normal makes equal intercepts.
$$\therefore \theta = 135^{\circ}$$
From Eq. (ii), we have
$$\dfrac {1}{2x - 1} = \tan 135^{\circ} = -1$$
$$\Rightarrow -2x + 1 = 1\Rightarrow x = 0$$
Then, from Eq. (i), $$y = 5$$
So, the required point is $$(0, 5)$$.