Tangents and its Equations Test 33

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Tangents and its Equations Test 33

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Weekly Quiz Competition

Question 1
1 / -0

Line $$y=x$$ and curve $$y=x^2+bx+c$$ touches at $$(1, 1)$$ then __________.

A
$$b=-1, c=1$$

B
$$b=1, c=2$$

C
$$b=1, c=1$$

D
$$b=0, c=1$$

Solution

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 $$c=1$$.
Question 2
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Equation of the normal of curve $$y=x^3-2x+4$$ at point $$(1, 3)$$ is __________.

A
$$x-y+4=0$$

B
$$x-y-4=0$$

C
$$x+y-4=0$$

D
$$x+y+4=0$$

Solution

$$y=x^3-2x+4$$
$$\therefore \dfrac{dy}{dx}=3x^2-2$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{(1.3)}=3(1)-2=3-2=1$$
$$\therefore$$ slope of tangent $$=1$$
$$\therefore$$ slope of normal $$=-1$$
$$\therefore$$ We get equation of Normal using
$$u-y_1=m(x-x_1)$$
$$\therefore y-3=-1(x-1)$$
$$\therefore y-3=-x+1$$
$$\therefore x+y-3-1=0$$
$$\therefore$$ Required equation of Normal is $$x+y-4=0$$
Question 3
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The equation to the tangent to $${\left( {{\dfrac xa}} \right)^n} + {\left( {{\dfrac yb}} \right)^n} = 2$$ at $$\left( {a,b} \right)$$

A
$$x+y=a+b$$

B
$${\dfrac xa} + {\dfrac yb} = 2$$

C
$${\dfrac xa} + {\dfrac yb} = 1$$

D
$${\dfrac xa} + {\dfrac yb} = 4$$

Solution

$${ (\dfrac { x }{ a } ) }^{ n }+{ (\dfrac { y }{ b } ) }^{ n }\\ n{ (\dfrac { x }{ a } ) }^{ n-1 }+n{ (\dfrac { y }{ b } ) }^{ n-1 }\dfrac { dy }{ dx } =0$$
At point $$(a,b)$$
$$x=a\ ,\ y=b$$
$$n+n\dfrac { dy }{ dx } =0\\ \dfrac { dy }{ dx } =-1=m\\ y=mx+c\\ y=-x+c$$
at$$(a,b)$$
$$b=-a+c$$
$$c=b+a$$
$$y+x=b+a$$
Question 4
1 / -0

Find the equations of the tangent and the normal to the curve $$x^2 +y^2=5$$ where the tangent is parallel to the lines $$x-y =0$$ respectively.

A
$$x-y=\pm \sqrt{10}, x+y=0$$

B
$$x-y=\pm {10}, x+y=0$$

C
$$x-y=\pm {10}, x+y=1$$

D
$$x-y=\pm \sqrt{10}, x+y=1$$

Solution

Let the tangent equation be $$x-y=c$$

Substituting above in circle equation ;

$$\Rightarrow (c+y) ^{2}+y^{2}=5$$

$$\Rightarrow 2y^{2}+2cy+c^{2}-5=0$$

As tangent intersects at only single point, above equation should have equal roots

$$\Rightarrow$$ Discriminant $$=0$$

$$\Rightarrow 4c^{2}=4×2×(c^{2}-5)$$

$$\Rightarrow c=\pm \sqrt {10}$$

$$\therefore $$ Equation of tangent is : $$x-y=\pm \sqrt {10}$$

As normal is perpendicular to tangent, let the equation of normal be $$x+y=k$$

And normal passes through the center of circle $$(0,0)$$

$$\Rightarrow 0+0=k$$

$$\Rightarrow k=0$$

$$\therefore$$ Equation of normal is $$x+y=0$$
Question 5
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The point on the curve $$y^{2}=x$$ where tangent makes $$45^{o}$$ angle with $$x-$$axis ?

A
$$(\dfrac{1}{2},\dfrac{1}{4})$$

B
$$(\dfrac{1}{4},\dfrac{1}{2})$$

C
$$(4,2)$$

D
$$(1,1)$$

Solution

$$ \dfrac{dy}{dx} = $$ 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 $$ \left(\dfrac{1}{4}, \dfrac{1}{2}\right) $$
Question 6
1 / -0

Equation of the normal to the curve $$y = -\sqrt{x} + 2$$ at the point of its interaction with the curve $$y = x$$ is

A
$$2x - y = 0$$

B
$$2x - y - 1 = 0$$

C
$$2x + y - 3 = 0$$

D
$$None\ of\ these$$

Solution

Equation of normal to the curve
$$y=-\sqrt{x}+2$$ at the point of intersection with $$y=x$$.
Point of intersection of
$$y=-\sqrt{x}+2$$, $$y=x$$
$$x=-\sqrt{x}+1$$
$$x-2=-\sqrt{x}$$
$$x^2-4x+4=x$$
$$x^2-5x+4=0$$
$$(x-4)(x-1)=0$$
$$x=4, 1$$
If $$x=4$$, $$(4, 4)$$ does not
Satisfy $$y=-\sqrt{x}+2$$
So, intersection point is $$(1, 1)$$
Normal at $$(1, 1)$$ to $$y=-\sqrt{x}+2$$
$$\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{x}}$$
$$\dfrac{dy}{dx}|_{(1, 1)}=\dfrac{-1}{2}$$
Slope of normal $$=\dfrac{(-1)}{\left(\dfrac{-1}{2}\right)}=2$$ and passes through $$(1, 1)$$
Equation of normal: $$2x-y=1$$.
Question 7
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The perpendicular distance from origin to the normal at any point to the curve $$x = a\left( {\cos \theta + \theta \sin \theta } \right)$$. $$y = a\left( {\sin \theta - \theta \cos \theta } \right)$$

A
$$a$$

B
$$a/2$$

C
$$a/3$$

D
$$2a$$

Solution

$$\dfrac{dy}{dx} = \dfrac{dy}{d\theta} \dfrac{d\theta}{dx} = tan \theta = \space slope\space of\space tangent$$

Slope of the normal $$=-\cot \space \theta$$ $$=\tan(90+\theta)$$

$$[y-a(\sin\theta - \theta \cos \theta)] = \dfrac{-\cos\theta}{\sin\theta} (x-a(\cos\theta+a \sin\theta))$$

$$x \cos\theta +y \sin\theta = a(1)$$

Now distance from $$(0,0)$$

$$d=\dfrac{(0+0-a)}{1} =a$$
Question 8
1 / -0

The slope of the tangent and normal to $$y={ x }^{ 2 }-3x+5$$ at $$(2,3)$$ are ______ and _____ respectively.

A
$$1,1$$

B
$$1,-1$$

C
$$-1,1$$

D
$$2,-2$$

Solution

the slope of the tangent at any point is the derivative of the equation at that point

therefore

$$\frac { d }{ dx } ({ x }^{ 2 }-3x+5)=2x-3$$

at$$(2,3)$$ $$\dfrac{dy}{dx}$$ is $$2(2)-3=1$$

therefore the slope of the tangent is $$1$$

normal is perpendicular to the tangent

therefore the slope of the normal*slope of the tangent=$$-1$$

therefore the slope of the normal is $$-1$$
Question 9
1 / -0

Find the angle between tangent of the curve $$y = (x + 1) (x - 3)$$ at the point where it cuts the axis of $$x$$.

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

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

C
$$\tan^{-1} 4$$

D
$$None\ of\ these$$

Solution

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 $$\tan^{-1}\left(\dfrac{8}{15}\right)$$.
Question 10
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If the tangent at the P of the curve $${y^2} = {x^3}$$ intersect the curve again at Q and the straight lines OP, OQ make angles $$\alpha ,\beta $$ with the x-axis, where O is the origin. then, $${{\tan \alpha } / {\tan \beta }}$$ has the value equal to

A
$$ - 1$$

B
$$ - 2$$

C
$$2$$

D
$$\sqrt 2 $$

Solution

Let the parametric coordinates at $$P$$ and $$Q$$ be :
$$(t_1^2, t_1^3), (t_2^2, t_2^3)$$
Then slope of the tangent at the point $$P\Rightarrow \dfrac{dy}{dx}=\dfrac{3x^2}{2y}=\dfrac{3}{2t_1}$$
Also, line joining $$P$$ and $$Q$$ would have a slope given by: $$\dfrac{t_2^3-t_1^3}{t_2^2-t_1^2}$$.
Equating both slopes, $$2\dfrac{t_2^2}{t_1^2}=\dfrac{t_2}{t_1}+1$$------(1)
Also. $$\tan \alpha=t_1, \tan \beta=t_2$$-------(2)
$$\therefore$$ From (1),
$$2\dfrac{\tan^2\beta}{\tan^2\alpha}=\dfrac {\tan \beta}{\tan \alpha}+1$$.
Let $$\dfrac{t_2}{t_1}=x$$
$$\therefore 2x^2=x+1$$
or, $$2x^2-x-1=0$$
or, $$2x^2-2x+x-1=0$$
or, $$2x(x-1)+1(x-1)=0\Rightarrow (x-1)(2x+1)=0$$
$$\therefore \boxed{x=1, 1/2}$$
i.e $$\dfrac{t_2}{t_1}=1\Rightarrow \boxed{t_2=t_1}$$ or, $$\dfrac{t_2}{t_1}=\dfrac{-1}{2}\Rightarrow \boxed{t_1=-2t_2}$$
$$\therefore$$ From (2) if $$t_2=t_1$$; and if $$t_1=2t_2$$
$$\tan \alpha=\tan \beta$$ $$\tan \alpha=-2\tan \beta$$
$$\therefore \boxed{\dfrac{\tan \alpha}{\tan \beta}=1}$$ $$\therefore \boxed{\dfrac{\tan \alpha}{\tan \beta}=-2}$$

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Tangents and its Equations Test 33

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Question 1

$$b=-1, c=1$$

$$b=1, c=2$$

$$b=1, c=1$$

$$b=0, c=1$$

Solution

Question 2

$$x-y+4=0$$

$$x-y-4=0$$

$$x+y-4=0$$

$$x+y+4=0$$

Solution

Question 3

$$x+y=a+b$$

$${\dfrac xa} + {\dfrac yb} = 2$$

$${\dfrac xa} + {\dfrac yb} = 1$$

$${\dfrac xa} + {\dfrac yb} = 4$$

Solution

Question 4

$$x-y=\pm \sqrt{10}, x+y=0$$

$$x-y=\pm {10}, x+y=0$$

$$x-y=\pm {10}, x+y=1$$

$$x-y=\pm \sqrt{10}, x+y=1$$

Solution

Question 5

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

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

$$(4,2)$$

$$(1,1)$$

Solution

Question 6

$$2x - y = 0$$

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

$$2x + y - 3 = 0$$

$$None\ of\ these$$

Solution

Question 7

$$a$$

$$a/2$$

$$a/3$$

$$2a$$

Solution

Question 8

$$1,1$$

$$1,-1$$

$$-1,1$$

$$2,-2$$

Solution

Question 9

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

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

$$\tan^{-1} 4$$

$$None\ of\ these$$

Solution

Question 10

$$ - 1$$

$$ - 2$$

$$2$$

$$\sqrt 2 $$

Solution

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