Tangents and its Equations Test 36

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Tangents and its Equations Test 36

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Question 1
1 / -0

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 ?

A
$$-\dfrac{5}{4}, -3, 3$$

B
$$0, \dfrac{1}{4},3$$

C
$$\dfrac{1}{4}, 0, 3$$

D
$$\dfrac{1}{4}, - \dfrac{1}{4}, 3$$

Solution

$$y=ax^{3}+bx^{2}+cx+8$$
$$\Rightarrow 0=-8a+4b+c(-2)+8$$
$$0=-8a+4b-2c+8 ----(1)$$

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

$$12a-4b=-3$$
$$-8a+4b=-2$$
____________
$$4a=-5$$
$$a=-5/4$$
$$b=-3$$
$$\boxed{\therefore a=-5/4, b=-3, c=3}$$
Question 2
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The equation of the normal to the curve $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ at the point $$(x_1, y_1)$$ on it is?

A
$$\dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2$$

B
$$\dfrac{a^x}{x_1}+\dfrac{b^2y}{y^1}=a^2+b^2$$

C
$$a^2xx_1+b^2yy_1=1$$

D
$$b^2xx_1-a^2yy_1=0$$

Solution

$$\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2} = 1$$
$$\cfrac{2x}{a^2} \cfrac{dx}{dy} + \cfrac{2y}{b^2} = 0$$
$$-\cfrac{dx}{dy} = \cfrac{a^2 y_1}{b^2 x_1}$$
Line - $$y = mx+c$$
$$y = \cfrac{a^2 y_1}{b^2 x_1} x +c$$
Putting $$(x_1,y_1)$$ in the line equation,we get
$$y_1 = \cfrac{a^2 y_1}{b^2} + c$$
$$c = y_1(1 - \cfrac{a^2}{b^2})$$
$$\cfrac{y}{y_1} = \cfrac{a^2 x}{b^2} + \cfrac{b^2-a^2}{b^2}$$
$$\cfrac{a^2 x}{x_1} -\cfrac{b^2 y}{y_1} = a^2 - b^2$$
Question 3
1 / -0

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:

A
$$\left( \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

B
$$\left( -\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 3 } \right) $$

C
$$\left( \cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } \right) $$

D
$$\left(- \cfrac { 1 }{ 4 } ,-\cfrac { 1 }{ 2 } \right) $$

Solution
Question 4
1 / -0

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

A
$$\dfrac{ab}{4}\sqrt{a^{2}+b^{2}}$$

B
$$4ab$$

C
$$\dfrac{b}{4a}({a^{2}+b^{2}})$$

D
$$none$$

Solution
Question 5
1 / -0

The numbers of tangent to the curve $$y - 2 = {x^5}$$ which are drawn
from point $$\left( {2,2} \right)$$ is/are

A
$$30$$

B
$$80$$

C
$$20$$

D
$$50$$

Solution

$$y-2=x^5 +2$$
$$y=x^5+2$$
$$y^1=5x^4$$
$$y^1=5(2)^4$$
$$y^1=5\times 16=80$$
Question 6
1 / -0

Equation of a normal to the curve $$y=x\log{x}$$, parallel to $$2x-2y+3=0$$ is

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

B
$$x-y=3e^{-2}$$

C
$$x-y=3e^{2}$$

D
$$x+y=3e^{2}$$

Solution
Question 7
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If the normal to the curve $$y=f\left( x \right)$$ at $${(3,4)}$$ makes angle $$\dfrac {3\pi}{4}$$ with $$\bar {OX}$$ then $$f^{ 1 }\left( 3 \right)=$$

A
$$-1$$

B
$$1$$

C
$${(-3/4)}$$

D
$${(4/3)}$$

Solution

$$\because y=f\left(x\right)$$
$${y}^{‘}={f}^{‘}\left(x\right)$$
Slope of tangent $$={f}^{‘}\left(x\right)$$
Slope of normal $$=-\dfrac{1}{{f}^{‘}\left(x\right)}$$
$$\Rightarrow \tan{\dfrac{3\pi}{4}}=-\dfrac{1}{{f}^{‘}\left(3\right)}$$
$$\Rightarrow -1=-\dfrac{1}{{f}^{‘}\left(3\right)}$$
$$\Rightarrow \boxed{{f}^{‘}\left(3\right)}=1$$
Question 8
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Equation of a tangent to the curve $$y=\cos(x+y),\ 0\le x\le 2\pi$$ that is parallel to the line $$x+2y=0$$ is

A
$$x+2y=\pi/2$$

B
$$x+2y=\pi/4$$

C
$$x+2y=\pi$$

D
$$x+y=\pi$$

Solution

Given curve y=cos (x+y)
we have find equation of tangent parallel to x+2y=0
slope of tangent is $$\dfrac{dy}{dx}$$
$$\dfrac{dy}{dx}=\dfrac{d}{dx} cos(x+y)=- \sin (x+y)\left(1+\dfrac{dy}{dx}\right)$$
$$\Rightarrow \dfrac{dy}{dx}(dx)(1+ \sin (x+y))=- \sin (x+y)$$
Now slope x+2y=0 is $$1+2\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{2}$$
Now $$+\dfrac{\sin (n+y)}{1+ \sin (x+y)}=+\dfrac{1}{2}$$
$$2 \sin (x+y)=1+ \sin (x+y)$$
$$\Rightarrow x+y=(2n+1)\dfrac{\pi }{2}$$ for n= positive integer
$$x=(2n+1)\dfrac{\pi }{2}-y$$
$$\therefore y=cos [(2n+1)\dfrac{\pi }{2}-y+y]= \cos \left[(2n+1)\dfrac{\pi }{2}\right]$$
$$\Rightarrow y=0$$
Now $$2x\leq x\leq 2x$$
putting n=0 $$x=\dfrac{\pi }{2}$$
n=-1 $$x=-\dfrac{3\pi }{2}$$
Thus i x= $$-\dfrac{3\pi }{2}$$ & $$x=\dfrac{\pi }{2}$$
$$\therefore $$ Points are $$\left(-\dfrac{3\pi }{2},0\right)$$ and $$\left(\dfrac{\pi }{2},0\right)$$
Now equation of tangent at $$\left(-\dfrac{3\pi }{2},0\right)$$ and $$\left(\dfrac{\pi }{2},0\right)$$ having solve $$-\dfrac{1}{2}$$ is
$$2x+4y+3\pi =0$$
and $$2x+4y-\pi =0$$
Question 9
1 / -0

The equation of normal to the curve $$x^{3}+y^{3}=8xy$$ at points where it is meet by the curve $$y^{2}=4x$$,other then origin is

A
$$y=x$$

B
$$y=-x+4$$

C
$$y=2x$$

D
$$y=-2x$$

Solution
Question 10
1 / -0

Tangents are drawn from a point on the circle $$x^2+y^2=25$$ to the ellipse $$9x^2+16y^2-144=0$$ then the angle between the tangents is

A
$$\frac{\pi}{4}$$

B
$$\frac{3\pi}{4}$$

C
$$\frac{\pi}{2}$$

D
$$\frac{2\pi}{3}$$

Solution