Tangents and its Equations Test 53

Result

Tangents and its Equations Test 53

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

For the curve $$x=t^2-1$$, $$y=t^2-t$$, the tangent is perpendicular to $$x$$-axis then

A
$$t=0$$

B
$$t=\dfrac{1}{2}$$

C
$$t=1$$

D
$$t=\dfrac{1}{\sqrt{3}}$$

Solution

Given curve is $$x=t^2-1, y=t^2-t$$

Derivating w.r.to $$t$$ we get

$$\dfrac{dx}{dt}=2t$$ -------(1)

and $$\dfrac{dy}{dt}=2t-1$$ -------(2)

dividing (2) by (1) we get

$$\dfrac{dy}{dx}=\dfrac{2t-1}{2t}$$

Therefore, the slope of the tangent is $$\dfrac{2t-1}{2t}$$

Given that the tangent is perpendicular to x-axis. Therefore, tangent is parallel to y-axis.

We know that slope of y-axis is infinity and the slopes of the two parallel lines are equal.

Therefore, slope of the tangent is infinity.

Hence, $$\dfrac{2t-1}{2t}=\dfrac{1}{0}$$

$$\implies 2t=0 \implies t=0$$
Question 2
1 / -0

Directions For Questions

Tangent at any point $$p_{1}$$ (other than $$(0, 0)$$) on the curve $$y = x^{3}$$ meets the curve again at $$p_{2}$$.
Tangent at $$p_{2}$$ meets the curve again at $$p_{3}$$ and so on.

...view full instructions

Abscissa of $$p_{1}, p_{2}, p_{3} .... p_{n}$$ are in

A
A.P.

B
G.P.

C
H.P

D
None

Solution
Question 3
1 / -0

If the line $$y=4x-5$$ touches to the curve $${ y }^{ 2 }=a{ x }^{ 3 }+b$$ at the point $$(2,3)$$ then $$7a+2b=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

The slope of given line $$y=4x-5$$ is $$m=4$$.

Now, since the given line touches the curve at $$(2,3)$$, the slope of curve $$\dfrac{dy}{dx}$$ at the given point is equal to the slope $$m$$ of the line.

Hence, for slope of curve, differentiating both sides w.r.t x:-

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

Putting the value $$x=2 $$ and $$y=3$$

$$6\dfrac{dy}{dx}=3a\times 4 \implies 6\times4=12a$$

$$\implies a=2$$

Now, the given point lies on the curve, so it must satisfy the equation of the curve:-

$$\implies (3)^2=2\times (2)^3+b$$

$$\implies b=-7$$

$$7a+2b=(7\times 2-2\times 7)=0$$

Hence, answer is option-(A).
Question 4
1 / -0

If the tangent at $$({x_1},{y_1})$$ to the curve $${x^3} + {y^3} = {a^3}$$ meets the curve again at $$({x_2},{y_2})$$, then

A
$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = - 1$$

B
$${{{x_2}} \over {{y_1}}} + {{{x_1}} \over {{y_2}}} = - 1$$

C
$${{{x_1}} \over {{x_2}}} + {{{y_1}} \over {{y_2}}} = - 1$$

D
$${{{x_2}} \over {{x_1}}} + {{{y_2}} \over {{y_1}}} = 1$$

Solution

Given curve equation is $$x^3+y^3=a^3$$.

The curve passes through the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Therefore, $$x_1^2+y_1^2=a^3$$ -------(1)

and $$x_2^3+y_2^3=a^3$$ -------(2)

subtracting (1) from (2) we get

$$y_2^3-y_1^3=-(x_2^3-x_1^3)$$ -------(3)

Therefore, $$3x^2+3y^2(\dfrac{dy}{dx})=0$$

$$\implies \dfrac{dy}{dx}=-\dfrac{x^2}{y^2}$$

Therefore, the slope of the tangent at $$(x_1,y_1)$$ is $$-\dfrac{x_1^2}{y_1^2}$$

The tangent passes through the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Therefore, the slope of the tangent is $$\dfrac{y_2-y_1}{x_2-x_1}$$

On comparing two slopes we get,

$$-\dfrac{x_1^2}{y_1^2}=\dfrac{y_2-y_1}{x_2-x_1}$$

we know that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

$$\implies -\dfrac{x_1^2}{y_1^2}=\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times \dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$

$$\implies -\dfrac{x_1^2}{y_1^2}=-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$

$$\implies x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_1^2$$

$$\implies x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$$

$$\implies (x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$$

$$\implies x-1y_2+x_2y_1=-x_1y_1$$

$$\implies \dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$$
Question 5
1 / -0

A point on the curve $$y = 2{x^3} + 13{x^2} + 5x + 9$$, the tangent at which passes through the origin is

A
$$(1, 15)$$

B
$$(1, -15)$$

C
$$(15, 1)$$

D
$$(-1, 15)$$

Solution

Let $$P(a,\,b)$$ be a point on the given curve $$y=2x^3+13x^2+5x+9$$
$$b=2a^3+13a^2+5a+9$$ --------- ( 1 )
Now, taking derivative of $$y$$.
$$\therefore$$ $$y'=6x^2+26x+5\Rightarrow y'\,at\,P=6a^2+26a+5$$.
Equation of tangent at $$P$$ is
$$y-b=(6a^2+26a+5)(x-a)$$
This tangent passes through origin means
$$b=a(6a^2+26a+5)$$
$$2a^3+13a^2+5a+9=6a^3+26a^2+5a$$
$$4a^3+13a^2-9=0$$
$$(a+1)(a+3)(4a-3)=0$$
Taking, $$a+1=0$$
We get, $$a=-1$$
Substituting $$a=-1$$ in equation (1) we get,
$$b=2(-1)^3+13(-1)^2+5(-1)+9$$
$$\therefore$$ $$b=15$$
$$\therefore$$ $$P(a,\,b)=(-1,\,15)$$
$$\therefore$$ A point on the curve $$y=2x^3+13x^2+5x+9$$, the tangent at which passes through the origin is $$(-1,\,15)$$.
Question 6
1 / -0

The curve that passes through the point $$(2,3)$$ and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

A
$${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad $$

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

C
$$y=\cfrac { 6 }{ x } $$

D
$${ x }^{ 2 }+{ y }^{ 2 }=13$$

Solution

Let $$B$$ be the pt. of contact $$y=f(x)$$ be the curve
then $$AB=BC$$ (given condition)
$$C\equiv (2x,0)$$ $$A\equiv (0,2y)$$
Slope $$m$$ of line $$AC$$ is
$$m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}$$
$$\dfrac{dy}{dx}=\dfrac{-y}{x}$$
$$\dfrac{dy}{y}=\dfrac{-dx}{x}$$
$$\ln y=-\ln x+c$$
$$yx=c$$
as $$(2, 3)$$ satisfies the curve
$$\therefore 2\times 3=c$$
$$\therefore y=\dfrac{6}{x}$$ is the curve
$$\therefore $$ Option $$C$$ is correct
Question 7
1 / -0

The tangent to the curve $$x = a(\theta - \sin \theta );y = a(1 + \cos \theta )$$ at the points $$\theta = (2k + 1)\pi ,k \in Z$$ are parallel to

A
$$y=x$$

B
$$y=-x$$

C
$$y=0$$

D
$$x=0$$

Solution

$$\Rightarrow x=a(\theta -\sin\theta ),\quad y=a(1+\cos\theta )\\ \Rightarrow \dfrac { dx }{ d\theta } =a(1-\cos\theta ),\quad \dfrac { dy }{ d\theta } =a(-\sin\theta )=-a\sin\theta \\ \therefore \dfrac { dy }{ dx } =\left( \dfrac { dy }{ d\theta } \right) \left( \dfrac { d\theta }{ dx } \right) =\dfrac { -a\sin\theta }{ a(1-\cos\theta ) } =\dfrac { -\sin\theta }{ 1-\cos\theta } $$
$$\therefore$$ Slope of tangent $$=\dfrac { -\sin\theta }{ 1-\cos\theta } $$
at $$\theta =(2k+1)\pi $$
Slope of tangent $$=\dfrac { -\sin(2k+1)\pi }{ 1-\cos(2k+1)\pi } =0$$
Since the slope of tangent $$=0$$, it will be parallel to $$x-$$ axis
Therefore $$y=0$$ is a possibility.
Question 8
1 / -0

A chord of parabola $${y}^{2}=4ax$$ subtends a right angle at the vertex. The tangents at the extremities of chord intersect on the line:

A
$$x+a=0$$

B
$$x+2a=0$$

C
$$x+3a=0$$

D
$$x+4a=0$$

Solution
Question 9
1 / -0

The slope of the tangent to the curve at a point $$(x,y) $$ on it is proportional to $$(x-2).$$ If the slope of the tangent to the curve at $$(10,-9)$$ on it is $$-3$$. The equation of the curves is .

A
$$y=k(x-2)^2$$

B
$$y=\dfrac{-3}{16}(x-2)^2+1$$

C
$$y=\dfrac{-3}{16}(x-2)^2+3$$

D
$$y=K(x+2)^2$$

Solution
Question 10
1 / -0

If $${p_1}$$ and $${p_2}$$ be the lengths of perpendiculars from the origin on the tangent and normal to the curve $${x^\frac{2}{3}} + {y^\frac{2}{3}} = {a^\frac{2}{3}}$$ respectively, then $$4{p_1}^2 + {p_2}^2 = $$

A
$$4a^2$$

B
$$2a^2$$

C
$$a^2$$

D
$$3{a^2}$$

Solution