MCQ Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 8

If the tangent to

$y^{2} = 4ax$ at the point

$(at^{2}, 2at)$ where

$|t| > 1$ is a normal to

$x^{2} - y^{2} = a^{2}$ at the point

$(a \sec \theta, a\tan \theta)$ , then

0%
$t = -cosec \theta$
0%
$t = -\sec \theta$
0%
$t = 2\tan \theta$
0%
$t = 2\cot \theta$

Explanation

We know equation of tangent is

$y=\dfrac{x}{t}+at$ and normal is

$xcos\theta+ycot\theta=2a$

So,

$(at^{2},2at)$ must satisfy the equation of normal

$\Rightarrow t^{2}cos\theta+2t\times cot\theta-2 =0$

So, the roots are

$t=-cosec\theta$ and

$t=2tan\theta$

The equation of the tangent to the curve

$y={ x }^{ 3 }-6x+5$ at

$(2,1)$ is

0%
$6x-y-11=0$
0%
$6x-y-13=0$
0%
$6x+y+11=0$
0%
$6x-y+11=0$

Explanation

The equation of the curve

$y={ x }^{ 3 }-6x+5$

$\Rightarrow \cfrac { dy }{ dx } =3{ x }^{ 2 }-6$

$\Rightarrow { \left( \cfrac { dy }{ dx } \right) }_{ 2,1 }=6$
Now, equation of the tangent at

$(2,1)$ is

$(y-1)=6(x-2)$

$\Rightarrow 6x-y-11=0$

The point on the curve

$y = \sqrt {x - 1}$ where the tangent is perpendicular to the line

$2x + y - 5 = 0$ is

0%
$(2, -1)$
0%
$(10, 3)$
0%
$(2, 1)$
0%
$(5, -2)$

Explanation

$\dfrac {dy}{dx} = \dfrac {1}{2\sqrt {x - 1}} = m_{1}$ is the slope of tangent to

$y=\sqrt{x-1}$
Slope of the line

$2x + y - 5 = 0$ is

$m_{2} = -2$
For lines are perpendicular

$m_{1} m_{2} = -1$

$\implies \left (\dfrac {1}{2\sqrt {x - 1}}\right )(-2) = -1$

$\implies \dfrac {2}{2\sqrt {x - 1}} = 1$

$\implies \sqrt {x - 1} = 1$
Squaring both sides,

$\implies x - 1 = 1$

$\implies x = 2$

$\therefore y = \sqrt {x - 1}$

$= \sqrt {2 - 1}$

$= \sqrt {1}$

$\therefore y = 1$

$\therefore (2, 1)$ is the point on the curve

$y=\sqrt{x-1}$

The slope of the normal to the curve

$x=1-a\sin { \theta }$ ,

$y=b\cos ^{ 2 }{ \theta }$ at

$\theta =\dfrac { \pi }{ 2 }$ is

0%
$\dfrac { a }{ 2b }$
0%
$\dfrac { 2a }{ b }$
0%
$\dfrac { a }{ b }$
0%
$\dfrac { -a }{ 2b }$

Explanation

Given,

$x=1-a\sin { \theta }$ and

$y=b\cos ^{ 2 }{ \theta }$
On differentiating with respect to

$\theta$ , we get

$\dfrac { dx }{ d\theta } =-a\cos { \theta }$
and

$\dfrac { dy }{ d\theta } =2b\cos { \theta } \left( -\sin { \theta } \right)$
Then,

$\dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\dfrac { 2b }{ a } \sin { \theta }$

$\therefore$ Slope of normal at the point

$\theta =\dfrac { \pi }{ 2 }$ is

$-\dfrac { dx }{ dy } =-\dfrac { 1 }{ { dy }/{ dx } }$

$=-\dfrac { 1 }{ \dfrac { 2b }{ a } \sin { \left( \dfrac { \pi }{ 2 } \right) } } =-\dfrac { a }{ 2b }$

If the straight line

$y -2x +1=0$ is the tangent to the curve

$xy+ax+by=0$ at

$x=1,$ then the values of

$a$ and

$b$ are respectively :

0%
1 and 2
0%
1 and -1
0%
-1 and 2
0%
-1 and -2
0%
1 and -2

Explanation

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.

The tangents to curve

$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$ which are parallel to straight line

$y=x$ , are

0%
$x+y=2$ and $x-y=\dfrac { 86 }{ 27 }$
0%
$x-y=2$ and $x-y=\dfrac { 86 }{ 27 }$
0%
$x-y=2$ and $x+y=\dfrac { 86 }{ 27 }$
0%
$x+y=2$ and $x+y=\dfrac { 86 }{ 27 }$

Explanation

Given,

$y={ x }^{ 3 }-2{ x }^{ 2 }+x-2$

On differentiating both sides with respect to

$x$ , we get

$\dfrac { dy }{ dx } =3{ x }^{ 2 }-4x+1$

and

$y=x$

$\Rightarrow \dfrac { dy }{ dx } =1$

$\therefore$ Slope of tangent will be

$3{ x }^{ 2 }-4x+1$ .

Since, the tangent is parallel to line

$y=x$ .

$\therefore 3{ x }^{ 2 }-4x+1=1$

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

$\Rightarrow x\left( 3x-4 \right) =0$

$\Rightarrow x=0,\dfrac { 4 }{ 3 }$

When

$x=0$ , then

$y=-2$

When

$x=\dfrac { 4 }{ 3 }$ , then

$y=\dfrac { -50 }{ 27 }$

Now, equation of tangents at point

$\left( 0,-2 \right)$ is

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

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

$\Rightarrow y+2=x$

$\Rightarrow x-y=2$ ...(i)

and equation of tangents at point

$\left( \dfrac { 4 }{ 3 } ,-\dfrac { 50 }{ 27 } \right)$ is

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

$y+\dfrac { 50 }{ 27 } =x-\dfrac { 4 }{ 3 }$

$\Rightarrow x-y=\dfrac { 50 }{ 27 } +\dfrac { 4 }{ 3 }$

$\Rightarrow x-y=\dfrac { 50+36 }{ 27 }$

$\Rightarrow x-y=\dfrac { 86 }{ 27 }$ ....(ii)

Hence, equations (i) and (ii) are required equations of the tangents.

The slope of tangent to the curve

$x=t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point

$(2, -1)$ is :

0%
$\dfrac {22}{7}$
0%
$\dfrac {6}{7}$
0%
$-6$
0%
None of these

Explanation

Given curves are

$x=t^{2} + 3t - 8$ ...

$(i)$ and

$y = 2t^{2} - 2t - 5$ ...

$(ii)$

$(2,-1),$ From

$(i)$

$t^{2}+3t-10=0\implies t=2$ or

$t=-5$

From

$(ii)$

$2t^{2}-2t-4=0\implies t^{2}-t-2=0\implies t=2$ or

$t=-1$

From both the solutions, we get

$t=2$

Differentiating both the equations w.r.t.

$t$ , we get

$\dfrac{dx}{dt}=2t+3$ ........

$(iii)$

$\dfrac{dy}{dt}=4t-2$ .......

$(iv)$

Now,

$\dfrac{dy}{dx} = \dfrac {\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$= \dfrac {4t-2}{2t+3}$ .... From

$(iii)$ and

$(iv)$

$\therefore \dfrac{dy}{dx} = \dfrac {4t-2}{2t+3}$ is the slope of tangent to the given curve

$\therefore \left|\dfrac{dy}{dx}\right|_{(2,-1)} = \left|\dfrac {4t-2}{2t+3}\right|_{t=2}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$ is the slope of tangent to the given curve at

$(2,-1)$

The points at which the tangent to the curve

$y = x^3 - 3x^2 - 9x + 7$ is parallel to the x-axis are

0%
$(3, - 20)$ and $(- 1, 12)$
0%
$(3, 20)$ and $(1, 12)$
0%
$(1, -10)$ and $(2, 6)$
0%
None of these

Explanation

Tangent to the curve is parallel to the axis is when slope of the tangent is 0.

$\therefore$ Equation of the curve is

$y=x^3-3x^2-9x+7=0$ ......

$(i)$

$\therefore \dfrac{dy}{dx}=3x^2-6x-9$

Now, the tangent is parallel to x-axis, then slope of the tangent is zero or we can say that

$\dfrac{dy}{dx}=0$ .

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow 3(x^2-2x-3)=0$

$\Rightarrow (x-3)(x+1)=0$

$\Rightarrow x=3, -1$

When

$x=3$ , then from Eq.

$(i),$ we get

$y=(3)^3-(3)\cdot (3)^2-9\cdot 3+7$

$=27-27-27+7=-20$

When

$x =-1,$ then from Eq.

$(i),$ we get

$y=(-1)^3-3(-1)^2-9(-1)+7$

$=-1-3+9+7=12$
Hence, the points at which the tangent is parallel to x-axis are

$(3, -20)$ and

$(-1, 12)$ .

The equation to the normal to the hyperbola

$\dfrac {x^{2}}{16} - \dfrac {y^{2}}{9} = 1$ at

$(-4, 0)$ is.

0%
$2x - 3y = 1$
0%
$x = 0$
0%
$x = 1$
0%
$y = 0$

Explanation

$\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$

Differentiating above equation, we get

$\dfrac { { 2x } }{ 16 } -\dfrac { 2{ y } }{ 9 } .\dfrac { dy }{ dx } =0$

$\implies \dfrac{dy}{dx}=\dfrac{9x}{16y}$

$\implies m=\left|\dfrac { dy }{ dx }\right|_{(-4,0)} =\dfrac{9(-4)}{ 16(0)} =\infty$ is the slope of tangent

Slope of the normal is

$-\dfrac{1}{m} =0$

Slope of the normal is zero so it is parallel to

$y$ -axis.

Equation of normal is

$y-0=0(x-(-4))$

$\therefore y=0$ is the equation of normal.

The slope of the tangent to the curve

$y=3{ x }^{ 2 }-5x+6$ at

$\left( 1,4 \right)$ is

0%
$-2$
0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

Given curve is

$y=3{ x }^{ 2 }-5x+6\Rightarrow \dfrac { dy }{ dx } =6x-5$

$\therefore$ Slope of tangent at

$\left( 1,4 \right) =6\times 1-5=1$

If the tangent at

$(1,1)$ on

${ y }^{ 2 }=x{ (2-x) }^{ 2 }$ meets the curve again at

$P$ , then

$P$ is

0%
$(4,4)$
0%
$(-1,2)$
0%
$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right)$
0%
$(1,2)$

Explanation

We have

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

$\Rightarrow 2y\cfrac { dy }{ dx } =3{ x }^{ 2 }-8x+4$

$\Rightarrow \cfrac { dy }{ dx } =\cfrac { 3{ x }^{ 2 }-8x+4 }{ 2y }$

$\Rightarrow { \left[ \cfrac { dy }{ dx } \right] }_{ (1,1) }=\cfrac { 3-8+4 }{ 2 } =-\cfrac { 1 }{ 2 }$

The equation of the tangent at

$(1,1)$ is

$-1=-\cfrac { 1 }{ 2 } (x-1)$

$\Rightarrow x+2y-3=0$ .....(ii)

On solving Eqs. (i) and (ii) we get

$x=9/4$ and

$y=3/8$

Hence,the coordinates of

$P$ are

$\left( \cfrac { 9 }{ 4 } ,\cfrac { 3 }{ 8 } \right)$ .

The tangent to the curve

$y=a{ x }^{ 2 }+bx$ at

$\left( 2,-8 \right)$ is parallel to

$X$ -axis. Then,

0%
$a=2, b=-2$
0%
$a=2, b=-4$
0%
$a=2, b=-8$
0%
$a=4, b=-4$

Explanation

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

$a=2,b=-8$

If the slope of the tangent to the curve

$y=a{ x }^{ 3 }+bx+4$ at

$(2,14) = 21$ , then the values of

$a$ and

$b$ are respectively

0%
$2,-3$
0%
$3,-2$
0%
$-3,-2$
0%
$2,3$

Explanation

$(2,14)$ lies on the curve

$y=ax^3+bx+4$

$8a+2b+4=14$

$4a+b=5$ ----(1)

$\dfrac{dy}{dx}=3ax^2+b|_{x=2}=21$

$12a+b=21$ --(2)

Solving equations (1) and (2), we get

$a=2, b=-3$

If the angle between the curves

$y = 2^x$ and

$y=3^x$ is

$\alpha,$ then the value of

$\tan \alpha$ is equal to :

0%
$\dfrac { \log \left( \dfrac {3}{2} \right) } { 1 + ( \log 2)( \log 3 ) }$
0%
$\dfrac {6}{7}$
0%
$\dfrac {1}{7}$
0%
$\dfrac { \log \left( 6 \right) } { 1 + ( \log 2)( \log 3 ) }$
0%
$0^o$

Explanation

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

$x=0$ ,

$\tan \alpha = \dfrac {3^0 \log 3 - 2^0 \log 2 }{1 + 3^0 \times 2^0 \log 2 \log 3 }$

$= \dfrac { \log \dfrac {3}{2} } { 1 + \log 2 \log 3 }$

If the tangent at each point of the curve

$y=\cfrac { 2 }{ 3 } { x }^{ 3 }-2a{ x }^{ 2 }+2x+5$ makes an acute angle with positive direction of X-axis then

0%
$a\ge 1$
0%
$-1\le a\le 1$
0%
$a\le -1$
0%
None of these

Explanation

On differentiating, We get

$\dfrac{dy}{dx}=2x^2-4ax+2$

If Tangent makes acute angle with +ve x axis, then

$\dfrac{dy}{dx}\ge0$ For all x

$x^2-2ax+1\ge0\\ \Rightarrow 4a^2-4\le0\Rightarrow a^2\le1\\ -1\le a\le1$

The equation of the tangent to the curve

$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$ at the point

$(x_{1}, y_{1})$ is

$\dfrac {x}{\sqrt {ax_{1}}} + \dfrac {y}{\sqrt {by_{1}}} = k$ . Then, the value of

$k$ is

0%
$2$
0%
$1$
0%
$3$
0%
$7$
0%
$\sqrt {2}$

Explanation

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,

$k = 1$

The slope of the normal to the curve

$y = x^2 - \dfrac{1}{x^2}$ at

$(-1, 0)$ is

0%
$\dfrac{1}{4}$
0%
$- \dfrac{1}{4}$
0%
$4$
0%
$-4$
0%
$0$

Explanation

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

$= - \dfrac{1}{dy/dx}$

$= - \dfrac{1}{-4} = \dfrac{1}{4}$

The point on the curve

$y = 5 + x - x^{2}$ at which the normal makes equal intercepts is

0%
$(1, 5)$
0%
$(0, -1)$
0%
$(-1, 3)$
0%
$(0, 3)$
0%
$(0, 5)$

Explanation

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

$(0, 5)$ .

A normal to parabola, whose inclination is

$30^o$ , cuts it again at an angle of.

0%
$\tan^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)$
0%
$\tan^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right)$
0%
$\displaystyle\tan^{-1}\cdot(2\sqrt{3})$
0%
$\displaystyle\tan^{-1}\left(\displaystyle\frac{1}{2\sqrt{3}}\right)$

Explanation

Equation of normal in terms of parameter

To parabola

$y^2=4ax$ at

$(at^2, 2at)$ , the equation of normal is

$y=-tx+2at+at^3$

To parabola

$x^2=4ay$ at

$(2at, at^2)$ the equation of normal is

$x+ty=2at+at^3$

Let the normal

$P(at_1,2at_1)$ by

$y= -t_1x + 2at_1+at_1^3$

$\theta$

$=-t_1=$ slope of normal

It meets the curve at

$\theta$ say

$(at_2^2,2at_2)$

Now angle between the normal and parabola

$=$ Angle the normal and tangent at

$\theta$ (i.e)

$t_2= x + at_2^2$

$\tan \phi$ =

$\dfrac{m_1-m_2}{1+m_1m_2}$

$=\dfrac{-t_1-\dfrac{1}{t_2}}{1+\dfrac{(-t_1)}{(-t_2)}}$

$=\dfrac{-(t_1t_2+1)}{t_2-t_1}$

$=\dfrac{-(t_1^2+1)}{t_2-t_1}$

$=\dfrac{-(-t_1^2-1)}{\dfrac{-2(1+t_1^2)}{(t_1)}}$

$=\dfrac{-t_1}{2}$

${\tan}\phi=\dfrac{\tan\theta}{2}$

hence,

$\phi$ =

$\tan^{-1}\dfrac{\tan30^.}{2} = \tan^{-1}(\dfrac{1}{2\times3^{\tfrac{1}{2}}})$

The slope of the tangent at the point

$(h, h)$ of the circle

$x^{2} + y^{2} = a^{2}$ is :

0%
$0$
0%
$1$
0%
$-1$
0%
Depends on $h$

Explanation

Slope of the tangent for the cicle

$x^{2}+y^{2}=a^{2}$ can be calculated by differentiating the circle's equation.

On differentiating w.r.t x,

$2x + 2y\dfrac{dy}{dx}=0$

Slope

$m$

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

Point given is

$(h,h)$

Putting point in the slope equation

$m=\dfrac{dy}{dx}=-\dfrac{h}{h}$

Slope

$m=-1$

Hence,

$(C)$

The angle between the curves

$x^{2} + y^{2} = 25$ and

$x^{2} + y^{2} - 2x + 3y - 43 = 0$ at

$(-3, 4)$ is

0%
$\tan^{-1}(1)$
0%
$\tan^{-1}\left (\dfrac {1}{68}\right )$
0%
$\dfrac {\pi}{2}$
0%
$\tan^{-1}\left (\dfrac {3}{4}\right )$

Explanation

slope of first curve at

$(-3,4)$

$m_1 = dy/dx = \dfrac{-x}{y} = \dfrac{3}{4}$

slope of second curve at

$(-3,4)$

$m_2 = \dfrac{-2x+2}{2y+3} = \dfrac{8}{11}$

angle between them

$= tan^{-1} \dfrac{m_1 - m_2}{1+m_1 +m_2}$

$=tan^{-1} \dfrac{1/44}{17/11} =tan^{-1} \dfrac{1}{68}$

The angle at which the curve

$y={ x }^{ 2 }$ and the curve

$x=\cfrac { 5 }{ 3 } \cos { t } ,y=\cfrac { 5 }{ 4 } \sin { t }$ intersect is

0%
$\tan ^{ -1 }{ \cfrac { 2 }{ 41 } }$
0%
$\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$
0%
$-\tan ^{ -1 }{ \cfrac { 2 }{ 41 } }$
0%
$2\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$

Explanation

Given

$y={ x }^{ 2 }...(i)$

$\quad x=\cfrac { 5 }{ 3 } \cos { t } ;y=\cfrac { 5 }{ 4 } \sin { t } ...(ii)$
Which is parametric equation, we change this equation is caresian equation as follows

$\cos { t } =\cfrac { 3 }{ 5 } x;\sin { t } =\cfrac { 4 }{ 5 } y$
On Squaring and adding both i.e.,

$\cos{t}$ and

$\sin {t}$ we get

$\cfrac { 9 }{ 25 } { x }^{ 2 }+\cfrac { 16 }{ 25 } { y }^{ 2 }=\cos ^{ 2 }{ t } +\sin ^{ 2 }{ t }$

$\Rightarrow 9{ x }^{ 2 }+16{ y }^{ 2 }=25....(iii)\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right]$

$\therefore$ The intersection points at Eq.(i) and (iii) are

$(1,1)$ and

$(-1,1)$
Now, slope of tangent of Eq.(i) at point

$(1,1)$ is

${ m }_{ 1 }=\cfrac { dy }{ dx } =2x\quad \therefore { m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=2$
And slope of tangent of Eq. (iii) at point

$(1,1)$ is

${ m }_{ 2 }=\cfrac { dy }{ dx } =-\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 1 }=\tan ^{ -1 }{ \left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$
similarly, slope of tangent of Eq. (i) at point

$(-1,1)$

${ m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (-1,1) }=-2$
And slope of tangent of Eq. (iii) at point

$(-1,1)$

${ m }_{ 2 }=\cfrac { dy }{ dx } =\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 2 }=\tan ^{ -1 }{ \left| \cfrac {-2- \cfrac { 9 }{ 16 } }{ 1-\cfrac { 18 }{ 16 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } \quad \quad$

The equation of the curve satisfying the differential equation

$y_{2}(x^{2} + 1) = 2xy_{1}$ passing through the point

$(0, 1)$ and having slope of tangent at

$x = 0$ as

$3$ is

0%
$y = x^{2} + 3x + 2$
0%
$y = x^{2} + 3x + 1$
0%
$y = x^{3} + 3x + 1$
0%
None of these

Explanation

Given

$\cfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \left( { x }^{ 2 }+1 \right) =2x\cfrac { dy }{ dx }$

$y={ x }^{ 2 }+3x+1$

Slope,

${ \left( \cfrac { dy }{ dx } \right) }_{ x=0 }={ (2x+3) }_{ x=0 }\Rightarrow 2\times 0+3\Rightarrow 3$

At point

$x=0$

$y={ (0) }^{ 2 }+3\times 0+1$

Thus this curve passes through point

$(0,1)$ and have slope of tangent at

$x=0$ as

$3$

$y={ x }^{ 2 }+3x+1$ is the answer

The slope of the tangent at each point of the curve is equal to the sum of the coordinate of the point. Then, the curve that passes through the origin is

0%
$x + y = e^{x} - 1$
0%
$e^{x} = x + y$
0%
$y = e^{x}$
0%
$y = e^{x} + 1$

Explanation

Given,

$\dfrac {dy}{dx}=x+y\Rightarrow dy=xdx+ydx\Rightarrow dy+dx=xdx+ydx+dx$

So,

$d(x+y)=(x+y+1)dx\Rightarrow dx=\dfrac {d(x+y+1)}{x+y+1}$

Integrating on both sides,

$x=\ln(x+y+1)$

On taking antilog,

$e^x=x+y+1$

Or.

$e^x-1=x+y$

So, Option A is correct answer.

A tangent PT is drawn to the circle

$x^2+y^2=4$ at the point

$P(\sqrt{3}, 1)$ . A straight line L, perpendicular to PT is a tangent to the circle

$(x-3)^2+y^2=1$ .

$(1)$ A possible equation of L is?

0%
$x-\sqrt{3}y=1$
0%
$x+\sqrt{3}y=1$
0%
$x-\sqrt{3}y=-1$
0%
$x+\sqrt{3}y=5$

Explanation

Slope of tangent to

$x^{2}+y^{2}=4$ at

$P(\sqrt{3},1)$ is given by differentiating given equation w.r.t. x

$\dfrac{dy}{dx}=\dfrac{-x}{y}=-\sqrt{3}$

So slope of line L

$=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

Let equation of L be

$x-\sqrt3y+c=0$

As L is tangent to

$(x-3)^{2}+y^{2}=1$

So perpendicular distance of L from its center should be equal to its radius i.e.

$1$ .

Centre

$(3,0)$

$\dfrac{|3+c|}{\sqrt4}=1$

$\Rightarrow c=-1 or -5$

So equation of L is

$x-\sqrt3y=1$

$x-\sqrt3y=5$

Hence option A is correct.

The points of the curve

$y={ x }^{ 3 }+x-2$ at which its tangent are parallel to the straight line

$y=4x-1$ are

0%
$\left( 2,7 \right) ,\left( -2,-11 \right)$
0%
$\left( 0,-2 \right) ,\left( { 2 }^{ 1/3 },{ 2 }^{ 1/3 } \right)$
0%
$\left( { -2 }^{ 1/3 },{ -2 }^{ 1/3 } \right) \left( 0,-4 \right)$
0%
$\left( 1,0 \right) ,\left( -1,-4 \right)$

Explanation

Given

$y={ x }^{ 3 }+x-2...(i)$

$y=4x-1....(ii)$

Slope of tangent to the curve (i)

$\cfrac { dy }{ dx } =3{ x }^{ 2 }+1$

slope of tangent at point

$\left( \alpha ,\beta \right)$ is

$\quad { \left| \cfrac { dy }{ dx } \right| }_{ \left( \alpha ,\beta \right) }^{ }=3{ \alpha }^{ 2 }+1...(iii)$

Given, tangent of curve (i) is parallel to line (ii)

$\therefore$ Slope of line (ii) is

$4$

$\therefore$ From Eq(iii) we get

$3{ \alpha }^{ 2 }+1=4\Rightarrow \alpha =\pm 1$

$\therefore \left( \alpha ,\beta \right)$ lie on curve (i)

$\beta ={ \left( \pm 1 \right) }^{ 2 }+\left( \pm 1 \right) -2\Rightarrow \beta =0,-4\quad$

$\therefore$ Points are

$(1,0)$ and

$(-1,-4)$

If the tangent at

$(1, 7)$ to the curve

$x^{2} = y - 6$ touches the circle

$x^{2} + y^{2} + 16x + 12y + c = 0$ then the value of

$c$ is

0%
$85$
0%
$95$
0%
$195$
0%
$185$

Explanation

Given equation of curve is

$x^{2}=y-6$

$\Rightarrow y=x^{2}+6$

$\Rightarrow \dfrac{dy}{dx}=2x$

Slope of tangent is

$m=\left|\dfrac{dy}{dx}\right|_{(1,7)}=2\times 1=2$

Equation of tangent at the point

$(1,7)$ is given by

$(y-7)=2(x-1)$

$\Rightarrow y-7=2x-2$

$\Rightarrow 2x-y+5=0$

Given equation of circle is

$x^{2} + y^{2} + 16x + 12y + c = 0$

$\Rightarrow (x+8)^{2}+(y+6)^{2}+c-64-36=0$

$\Rightarrow (x+8)^{2}+(y+6)^{2}=100-c$

If the tangent touches the circle, then

Distance from centre

$=$ radius

$\Rightarrow$ Distance of

$(-8,-6)$ from

$2x-y+5=0$ is the radius

$d=\left|\dfrac{2(-8)-(-6)}{\sqrt{4+1}}\right|=\left|\sqrt{5}\right|$

$\Rightarrow Radius=\sqrt{100-c}=\sqrt{5}$

$\Rightarrow c={95}$

The point on the curve

$y^{2}=x$ where tangent makes

$45^{o}$ angle with

$x-$ axis ?

0%
$(\dfrac{1}{2},\dfrac{1}{4})$
0%
$(\dfrac{1}{4},\dfrac{1}{2})$
0%
$(4,2)$
0%
$(1,1)$

Explanation

$\dfrac{dy}{dx} =$ Slope of the tangent at that point of the curve.

Given curve is

$x = y^2$

Keep

$y = a, \, then \, x = y^2$

point

$(a^2, a)$

$\displaystyle \dfrac{dy}{dx} (y^2 - x) = 0 \dfrac{dy}{dx} = \dfrac{1}{2y} = \dfrac{1}{2a} = \tan(45^0)$

$\Rightarrow a= \dfrac{1}{2} \quad a^2= \dfrac{1}{4}$

point

$\left(\dfrac{1}{4}, \dfrac{1}{2}\right)$

The tangent to

$\left( a{ t }^{ 2 },2at \right)$ is perpendicular to X-axis at _____ point

$t\in R$ .

0%
$(4a,4a)$
0%
$(a,2a)$
0%
$(0,0)$
0%
$(a,-2a)$

Explanation

Here

$\left( x,y \right) =\left( a{ t }^{ 2 },2at \right)$

$\therefore x=a{ t }^{ 2 };y=2at$

$\therefore \cfrac { dx }{ dt } =2at;\cfrac { dy }{ dt } =2a$

$\therefore \cfrac { dy }{ dx } \cfrac { dy/dt }{ dx/dt } =\cfrac { 2a }{ 2at } =\cfrac { 1 }{ t }$
Since, tangent is perpendicular to X-axis slope is undefined

$\therefore$

$t=0$

$\therefore t=0$

$\therefore \left( a{ t }^{ 2 },2at \right) =\left( 0,0 \right)$
Thus, tangent is perpendicular to X-axis at

$(0,0)$

If the tangent to the curve

$y=x\log { x }$ at

$\left( c,f\left( x \right) \right)$ is parallel to the line-segment joining

$A\left(1,0\right)$ and

$B\left(e,e\right)$ , then c=...... .

0%
$\dfrac {e-1}{e}$
0%
$\log { \dfrac { e-1 }{ e } }$
0%
${ e }^{ \dfrac { 1 }{ 1-e } }$
0%
${ e }^{ \dfrac { 1 }{ e-1 } }$

Explanation

slope of AB

$\displaystyle =\frac{e}{e-1}$

$\displaystyle \frac{dy}{dx}=1+\log x$

slope at

$(c,f\left(x\right ))$

$\displaystyle 1+\log c=\frac{e}{e-1}\implies=e^{\frac{1}{e-1}}$

The equation of the curve passing through

$(1,3)$ whose slope at any point

$(x,y)$ on it is

$\dfrac { y }{ { x }^{ 2 } }$ is given by

0%
$y={ 3e }^{ -1/x }$
0%
$y={ 3e }^{ 1-1/x }$
0%
$y={ ce }^{ 1/x }$
0%
$y={ 3e }^{ 1/x }$

Explanation

$\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}\,\,\,\,\,\,\,\left( {given} \right)$

$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{{dx}}{{{x^2}}}} }$

$\Rightarrow \log y = - \dfrac{1}{x} + c$

$\Rightarrow y = {e^{ - \frac{1}{x} + c}}$

$\Rightarrow y = {e^{ - \frac{1}{x}}},c$

since this curve is passing through point $\left( {1,3} \right)$

so, $3 = {e^{ - \dfrac{1}{t}}}.c$

$\Rightarrow c = 3e$

therefore , eqn of the curve is

$y = {e^{ - \dfrac{1}{x}}}.3e$

$\Rightarrow y = 3{e^{ - \dfrac{1}{x} + 1}}$

$\Rightarrow y = 3{e^{1 - \dfrac{1}{x}}}$

Line

$y=x$ and curve

$y=x^2+bx+c$ touches at

$(1, 1)$ then __________.

0%
$b=-1, c=1$
0%
$b=1, c=2$
0%
$b=1, c=1$
0%
$b=0, c=1$

Explanation

Given line

$l : y =x$

$l : x-y=0$

$\therefore$ slope of line

$l=1$
Now

$y=x^2+bx+c$

$\therefore \dfrac{dy}{dx}=2x+b$

$\therefore \left(\dfrac{dy}{dx}\right)_{P(1, 1,)}=2(1)+b$

$=2+b$
But slope of tangent of the curve

$=1$ .

$\therefore 2+b=1$

$\therefore b=-1$
Now

$(1, 1)\in y=x^2+bx+c$

$\therefore 1=1+b(1)+c$

$\therefore b+c=0$

$\therefore c=-b$

$c=-(-1)$

$\therefore c=1$

$\therefore$ We have

$b=-1$ and

$c=1$ .

Find the angle between tangent of the curve

$y = (x + 1) (x - 3)$ at the point where it cuts the axis of

$x$ .

0%
$\tan^{-1} \left(\dfrac{8}{15}\right)$
0%
$\tan^{-1} \left(\dfrac{15}{8}\right)$
0%
$\tan^{-1} 4$
0%
$None\ of\ these$

Explanation

The curve

$y=(x+1)(x-3)=x^2-2x-3$ .........(1) cuts the axis of

$x$ at

$(-1,0)$ and

$(3.0)$ . These points are obtained by putting

$y=0$ in the equation (1).

Now the slope of the tangent to the curve (1) at

$(-1,0)$ be

$(m_1)=\left.\dfrac{dy}{dx}\right|_{(-1,0)}=[2x-2]_{(-1,0)}=-4$ .

Again the slope of the tangent to the curve (1) at

$(-1,0)$ be

$(m_2)=\left.\dfrac{dy}{dx}\right|_{(3,0)}=[2x-2]_{(3,0)}=4$ .

Now the angle between the tangents to (1) at those points be

$\tan^{-1}\left(\dfrac{m_1-m_2}{1+m_1.m_2}\right)$

$=\tan^{-1}\left(\dfrac{8}{15}\right)$ .

So the required angle be

$\tan^{-1}\left(\dfrac{8}{15}\right)$ .

If the curves

$y^2 = 4ax$ and

$xy = c^2$ cut orthogonally then

$\dfrac{c^4}{a^4} =$

0%
$4$
0%
$8$
0%
$16$
0%
$32$

Explanation

$y^2=4ax, xy=c^2$ cut orthogonally

Let they intersect at

$(x_1, y_1)$

$y^2=4ax$

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

$\dfrac{dy}{dx}=\dfrac{2a}{y}$

$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{2a}{y_1}$ …..

$(1)$

$xy=c^2$

$y+x\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}|_{(x_1, y_1)}=\dfrac{-y_1}{x_1}$ ……

$(2)$

From

$(1), (2)$

$(\because m_1m_2=-1)$

$\dfrac{2a}{\not{y_1}}\times \dfrac{\not{-}\not{y_1}}{x_1}=\not{-}1$

$x_1=2a$

From

$y^2=4ax$

$y_1=\sqrt{8a^2}=2\sqrt{2}a$

$xy=c^2$

$x_1y_1=c^2$

$2a(2\sqrt{2}a)=c^2$

$\dfrac{c^2}{a^2}=4\sqrt{2}$

$\dfrac{c^4}{a^4}=32$ .

An equation of the tangent to the curve

$y=x^{4}$ from the point

$(2,0)$ not on the curve is:

0%
$y=0$
0%
$x=0$
0%
$x+y=0$
0%
$none\ of\ these$

Explanation

let a,b be point on curve (x, y)

$\displaystyle b = a^4$

$\displaystyle \frac{dy}{dx} (x^4) = 4x^3 = 4a^3$

$\displaystyle (y- y_1) = m (x - x_1) \rightarrow$ be tangent equation passes through (2, 0)

$\displaystyle (b-0) = 4a^3 (a-2)$

$\displaystyle a^4 = 4a^3 (a-2)$

$\displaystyle 3a^4 - 8a^3 = 0$

$\displaystyle a^3 (3a - 8) = 0$

$\displaystyle a= 0 \quad or \quad a= \frac{8}{3} \quad (0,0)$ &

$\left[\dfrac{8}{3},\left(\dfrac{8}{3} \right)^4\right]$ points on curve

$\displaystyle b=0 \quad or \quad b= \left(\frac{8}{3}\right)^4$

$\displaystyle (y - 0) = \frac{0-0}{2-0} (x-0)\Rightarrow y = 0 \rightarrow$ tangent (1)

$\displaystyle \left[y-\left(\frac{8}{3}\right)^4\right] = \frac{(\frac{8}{3})^4_{-0}}{(\frac{8}{3})_{-2}} \quad (x - \frac{8}{3}) \rightarrow$ tangent (2)

1058172_1038190_ans_b5c2db2c6f76484f88ed313c09fb571a.jpg

A point P moves such that sum of the slopes of the normal drawn from it to the hyperbola

$xy=16$ is equal to the sum of the ordinates of the feet of the normal. Let 'P' lies on the curve C, then.
The equation of 'C' is?

0%
$x^2=4y$
0%
$x^2=16y$
0%
$x^2=12y$
0%
$y^2=8x$

Explanation

Let P(h,k) be the point in the plane of hyperbola

$xy=16=4^2$

The equation of the normal at a point

$(4t, \dfrac {4} {t})$ to the hyperbola

$xy=4^2$ is

$xt^3-yt-4t^4+4=0$

if if passes through

$P(h,k)$ then,

$ht^3-yt-4t^4+4=0$

$\Rightarrow 4t^4-ht^3+kt-4=0$

this is a fourth degree equation in t. So, it gives 4 values of t,

$t_1,t_2,t_3,t_4$ corresponding to each value of t there is a point on hyperbola such that the normal passes through

$P(h,k)$ .

let the 4 points be

$A(ct_1, \dfrac{c}{t_1})$ ,

$B(ct_2, \dfrac{c}{t_2})$ ,

$C(ct_3, \dfrac{c}{t_3})$ and

$D(ct_4, \dfrac{c}{t_4})$

such that normal at these points pass through

$P(h,k)$ .

Since

$t_1,t_2,t_3,t_4$ are roots of equation (1).

Therefore,

$t_1+t_2+t_3+t_4 = \dfrac{h}{c}=\dfrac{h}{4}$ .......(2)

$\sum t_1t_2 =0$ .......(3)

$\sum t_1t_2t_3 = - \dfrac{k}{4}$ .........(4)

$t_1t_2t_3t_4 =-1$ .......(5)

It s given that the sum of the slopes of the normals at A,B,C and D is equal to the sum of the coordinates of these points.

Therefore

$t_1^2+t_2^2+t_3^2+t_4^2 = \dfrac{c}{t_1}+\dfrac{c}{t_3}+\dfrac{c}{t_3}+\dfrac{c}{t_4}=\dfrac{4}{t_1}+\dfrac{4}{t_2}+\dfrac{4}{t_3}+\dfrac{4}{t_4}$

$(t_1+t_2+t_3+t_4 )^2-2\sum t_1t_2=4\left ( \dfrac {\sum t_1t_2t_3}{t_1t_2t_3t_4} \right )$

$\Rightarrow \dfrac{h^2}{4}-2\times 0 = 4 \times \left ( \dfrac{\dfrac {-k}{4}} {-1} \right )$

$\Rightarrow \dfrac{h^2}{4} = k$

$\Rightarrow h^2 = 16k$

so the locus of

$(h,k)$ is

$x^2=16y$ , which is a parabola

The equation of the tangent to the curve

$y = b{e^{ -\dfrac{x}{a}}}$ at a point , where

$x=0$ is

0%
$\dfrac{x}{a} - \dfrac{y}{b} = 1$
0%
$\dfrac{y}{b} - \dfrac{x}{a} = 1$
0%
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
0%
$\dfrac{x}{b} + \dfrac{y}{a} = 1$

Explanation

$y=b{ e }^{ \cfrac { -x }{ a } }$ when

$x=0\, y=b$

$\dfrac { dy }{ dx } =\cfrac { -b }{ a } { e }^{ \cfrac { -x }{ a } }$ at

$x=0$ ,

$\dfrac { dy }{ dx } =\cfrac { -b }{ a }$

$y=\cfrac { -bx }{ a } +c$

$y=\cfrac { -bx }{ a } +b\\ \cfrac { y }{ b } =\cfrac { -x }{ a } +1\\ \cfrac { x }{ a } +\cfrac { y }{ b } =1$

Area of the triangle formed by the tangent, normal to the curve

$x^{2}/a^{2}+y^{2}/b^{2}=1$ at the point

$(a/\sqrt{2} , b/\sqrt{2})$ and the

$x-$ axis is

0%
$\dfrac{ab}{4}\sqrt{a^{2}+b^{2}}$
0%
$4ab$
0%
$\dfrac{b}{4a}({a^{2}+b^{2}})$
0%
$none$

Find the slope of the normal to the curve

$2x^{2} - xy + 3y^{2} = 18$ at

$(3,1)$ .

0%
$\dfrac {-11}{3}$
0%
$\dfrac {3}{11}$
0%
$\dfrac {11}{3}$
0%
$\dfrac {-3}{11}$

Explanation

Equation of curve

$2{ x }^{ 2 }-xy+3{ y }^{ 2 }=18$

Differentiating above equation w.r.t

$x,$

$4x-y-x\dfrac { dy }{ dx }+6y\dfrac { dy }{ dx } =0$

$4x-y+\dfrac { dy }{ dx } (6y-x)=0$

$\dfrac { dy }{ dx } (6y-x)=y-4x$

$\dfrac { dy }{ dx } =\dfrac { y-4x }{ 6y-x }$

$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { 1-4\left( 3 \right) }{ 6-3 }$

$\dfrac { dy }{ dx| } _{ (3,1) }=\dfrac { -11 }{ 3 }$

Slope of normal to the curve

$=\dfrac { -1 }{ \dfrac { dy }{ dx| } _{ (3,1) } }$

$=\dfrac { -1 }{ \dfrac { -11 }{ 3 } }$

$=\dfrac { 3 }{ 11 }$

$\boxed { \therefore Ans =\dfrac { 3 }{ 11 } }$

Area of the triangle formed by the tangent at

$x=2$ on the curve

$y= \dfrac{8}{4+x^2}$ with the coordinate axes is (in sq. units)

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

$y=\dfrac {8}{4+x^2}$

$At \,\,x=2\ y=1$

$y' =\dfrac {-8(2x)}{(4+x^2)^2}=\dfrac {-16\times 2}{8^2}=\dfrac {-1}{2}$

Tangent is

$y=mx+c$

$y=\dfrac {-1}{2}x+c$

passes through

$(2, 1)\ C=2$

$2y+x=4$

Base of triangle

$=4$

Height of triangle

$=\dfrac {4}{2}=2$

Area of

$\Delta =\dfrac {1}{2}\times 4\times 2=4\ sq.unit$

1416924_1090027_ans_25bd3feb84ce43e795654362a41fb34a.png

Find the points on the ellipse

$\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{9}=1$ , on which the normals are parallel to the line

$3x-y=1$ .

0%
$(\pm\dfrac{6}{\sqrt {10}},\pm\dfrac{3}{\sqrt {10}})$
0%
$(\pm\dfrac{3}{\sqrt {10}},\pm\dfrac{1}{\sqrt {10}})$
0%
$(\pm\dfrac{1}{\sqrt {10}},\pm\dfrac{2}{\sqrt {10}})$
0%
None of these

The curve

$y = ax^3 + bx^2 + cx + 8$ touches

$x-$ axis at

$P(-2, 0)$ and cuts

$y-$ axis at a point

$Q$ where its gradient is

$3$ . The values of

$a, b, c$ are respectively ?

0%
$-\dfrac{5}{4}, -3, 3$
0%
$0, \dfrac{1}{4},3$
0%
$\dfrac{1}{4}, 0, 3$
0%
$\dfrac{1}{4}, - \dfrac{1}{4}, 3$

Explanation

$y=ax^{3}+bx^{2}+cx+8$

$\Rightarrow 0=-8a+4b+c(-2)+8$

$0=-8a+4b-2c+8 ----(1)$

It touches

$y-$ axis at

$Q\Rightarrow x=0$

$y=8$

$Q.(0,8)$

$\dfrac{dy}{dx}=3ax^{2}+2bx+c$

Given gradient is

$3$ at

$G(0,8)$

$3=c$

$\therefore c=3$

$\because$ the curve touches

$x-$ axis

$\therefore$ slope of the tangent at

$P$ is

$0$

$\dfrac{dy}{dx}=3ax^{2}+2bx+3$

$0=12a-4b+3$

$12a-4b=-3 ---- (1)$

eqn

$(1)$

$-8a+4b-6+8=0$

$-8a+4b=-2 --- (ii)$

Adding

$(10$ &

$(2)$ , we get

$12a-4b=-3$

$-8a+4b=-2$

____________

$4a=-5$

$a=-5/4$

$b=-3$

$\boxed{\therefore a=-5/4, b=-3, c=3}$

Number of different points on the curve

$y^2=x(x+1)^2$ where the tangent to the curve drawn at

$(1, 2)$ meets the curve, is?

0%
$0$
0%
$1$
0%
$2$
0%
$3$

The equation of tangent to the curve

$\sqrt {x}+ \sqrt {y}= \sqrt {a}$ at the point

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ is-

0%
$\dfrac { x }{ \sqrt { { x }_{ 1 } } } +\dfrac { y }{ \sqrt { { y }_{ 1 } } } =\dfrac { 1 }{ \sqrt { a } }$
0%
$\dfrac { x }{ \sqrt { { x }_{ 1 } } } +\dfrac { y }{ \sqrt { { y }_{ 1 } } } ={ \sqrt { a } }$
0%
$x\sqrt { { x }_{ 1 } } +y\sqrt { { y }_{ 1 } } =\sqrt { a }$
0%
$None\ of\ these$

Explanation

We have,

$\sqrt{x}+\sqrt{y}=\sqrt{a}$

Differentiation this equation with respect to $x$ and we get,

$\dfrac{d}{dx}\left( \sqrt{x}+\sqrt{y} \right)=\dfrac{d}{dx}\sqrt{a}$

$\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=0$

$\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}}$

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

At point $\left( {{x}_{1}},{{y}_{1}} \right)$

${{\left( \dfrac{dy}{dx} \right)}_{\left( {{x}_{1}},{{y}_{1}} \right)}}=-\dfrac{\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}}$

We know that,

Equation of tangent.

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

At point $\left( {{x}_{1}},{{y}_{1}} \right)$

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

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

$y\sqrt{{{x}_{1}}}-{{y}_{1}}\sqrt{{{x}_{1}}}=-x\sqrt{{{y}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}$

$y\sqrt{{{x}_{1}}}+x\sqrt{{{y}_{1}}}={{y}_{1}}\sqrt{{{x}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}$

On divide $\sqrt{{{x}_{1}}{{y}_{1}}}$ both side and we get,

$\dfrac{y\sqrt{{{x}_{1}}}+x\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}\sqrt{{{y}_{1}}}}=\dfrac{{{y}_{1}}\sqrt{{{x}_{1}}}+{{x}_{1}}\sqrt{{{y}_{1}}}}{\sqrt{{{x}_{1}}}\sqrt{{{y}_{1}}}}$

$\dfrac{y}{\sqrt{{{y}_{1}}}}+\dfrac{x}{\sqrt{{{x}_{1}}}}=\sqrt{{{y}_{1}}}+\sqrt{{{x}_{1}}}$

$\dfrac{x}{\sqrt{{{x}_{1}}}}+\dfrac{y}{\sqrt{{{y}_{1}}}}=\sqrt{a}$

Hence, this is the answer.

The equation normal to the curve

$x^{2/3}+y^{2/3}=a^{2/3}$ at the point

$(a, 0)$ is-

0%
$x=a$
0%
$x=-a$
0%
$y=a$
0%
$y=-a$

Explanation

We have,

${{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}}={{a}^{\frac{2}{3}}}$

Then,

$\dfrac{d}{dx}\left( {{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}} \right)=\dfrac{d}{dx}\left( {{a}^{\frac{2}{3}}} \right)$

$\dfrac{2}{3}{{x}^{-\frac{1}{3}}}+\dfrac{2}{3}{{y}^{-\frac{1}{3}}}\dfrac{dy}{dx}=0$

${{x}^{-\frac{1}{3}}}+{{y}^{-\frac{1}{3}}}\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}={{\left( -\dfrac{x}{y} \right)}^{-\frac{1}{3}}}$

$\dfrac{dy}{dx}={{\left( -\dfrac{y}{x} \right)}^{\frac{1}{3}}}$

At point $\left( a, 0 \right)$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}={{\left( -\dfrac{y}{x} \right)}^{\frac{1}{3}}}_{\left( a,0 \right)}$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}={{\left( -\dfrac{0}{a} \right)}^{\frac{1}{3}}}$

${{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}=0$

Equation of normal is

$y-{{y}_{1}}=-\dfrac{1}{{{\left( \dfrac{dy}{dx} \right)}_{\left( a,0 \right)}}}\left( x-{{x}_{1}} \right)$

$y-{{y}_{1}}=\dfrac{1}{0}\left( x-{{x}_{1}} \right)$

Equation of normal at point $\left( a,0 \right)$ and we get,

$y-0=\dfrac{1}{0}\left( x-a \right)$

$x-a=0$

$x=a$

Hence, this is the answer.

The numbers of tangent to the curve

$y - 2 = {x^5}$ which are drawn
from point

$\left( {2,2} \right)$ is/are

0%
$30$
0%
$80$
0%
$20$
0%
$50$

Explanation

$y-2=x^5 +2$

$y=x^5+2$

$y^1=5x^4$

$y^1=5(2)^4$

$y^1=5\times 16=80$

The curve y

$=ax^3+bx^2+cx+5$ touches the x-axis at

$P(-2, 0)$ then

$C=?$

0%
$4a+5$
0%
$4a-5$
0%
$5-4a$
0%
$0$

Explanation

As the curve pases through the point

$(-2,0)$

So,

$0 = -8a+4b-2c+5$

$8a-4b+2c = 5.....(i)$

Also, the tangent parallel to x-axis is 0. Hence,

$3ax^2+2bx+c = \cfrac{dy}{dx}$

$12a-4b+c = 0.......(ii)$

Subtracting (ii) from (i), we get

$-4a+c = 5$

$c = 4a+5$

If the normal to the curve

$y=f\left( x \right)$ at

${(3,4)}$ makes angle

$\dfrac {3\pi}{4}$ with

$\bar {OX}$ then

$f^{ 1 }\left( 3 \right)=$

0%
$-1$
0%
$1$
0%
${(-3/4)}$
0%
${(4/3)}$

Explanation

$\because y=f\left(x\right)$

${y}^{‘}={f}^{‘}\left(x\right)$

Slope of tangent

$={f}^{‘}\left(x\right)$

Slope of normal

$=-\dfrac{1}{{f}^{‘}\left(x\right)}$

$\Rightarrow \tan{\dfrac{3\pi}{4}}=-\dfrac{1}{{f}^{‘}\left(3\right)}$

$\Rightarrow -1=-\dfrac{1}{{f}^{‘}\left(3\right)}$

$\Rightarrow \boxed{{f}^{‘}\left(3\right)}=1$

The curve satisfying D.E

$y dx - (x+3{y}^{2})dy=0$ and passing through the point

$(1,1)$ also passes through the point:

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$\left( \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right)$
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$\left( -\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right)$
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$\left( \cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } \right)$
0%
$\left(- \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right)$

If the curves

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1$ and

$y^{2}=16x$ intersect at right angles then value of

$a^{2}$ is

0%
$2$
0%
$4$
0%
$\dfrac{8}{3}$
0%
$\dfrac{16}{3}$

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 8 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 8 - MCQExams.com

0:0:2

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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