Self Studies
Self Studies
Self Studies
Self Studies

Conic Sections Test - 52

Result Self Studies

Conic Sections Test - 52
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

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.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The length of latus rectum of the parabola whose parametric equations are $$ x = t^{2} + t + 1$$, $$y = t^{2}-  t + 1$$, where $$t \in R$$, is equal to?
    Solution
    $$x=t^2+t+1$$
    $$y=t^2-t+1$$
    Adding above equations, we get
    $$\therefore t=\dfrac{x-y}{2}$$
    Substituting this in first equation, we get
    $$\therefore x=\left(\dfrac{x-y}{2}\right)^2+\dfrac{x-y}{2}+1$$
    $$\therefore \dfrac{x+y-2}{2}=\left(\dfrac{x-y}{2}\right)^2$$
    $$\therefore (x-y)^2=2(x+y-2)$$
    So, length of latus rectum = 2
  • Question 2
    1 / -0
    For the variable, the locus of the point of intersection of the lines $$3tx-2y+6t=0$$ and $$3x+2ty-6=0$$ is
    Solution
    Given equation of lines are
    $$3tx-2y+6t=0......(i)$$
    and 
    $$3x+2ty-6=0.....(ii)$$

    On multiplying Eq $$(i)$$ by $$t$$ and then adding in eq.$$(ii)$$, we get
    $$(3{t}^{2}+3)x+6{t}^{2}-6=0$$
    $$\Rightarrow$$ $$x=\cfrac{2(-1{t}^{2})}{(1+{t}^{2})}$$
    $$\Rightarrow$$ $$x+x{t}^{2}=2-2{t}^{2}$$
    $$\Rightarrow$$ $$(x+2){t}^{2}=(2-x)$$
    $$\Rightarrow$$ $${t}^{2}=\cfrac{2-x}{2+x}.....(iii)$$

    On multiplying eq $$(ii)$$ by $$t$$ and then subtract from eq $$(i)$$ we get
    $$(-2-2{t}^{2})y+6t+6t=0$$
    $$12t=2(1+{t}^{2})y$$

    On squaring both sides, we get
    $$144{t}^{2}=4{y}^{2}{(1+{t}^{2})}^{2}$$
    $$\Rightarrow$$ $$144\left(\cfrac{2-x}{2+x}\right)=4{y}^{2}{\left(1+\cfrac{2-x}{2+x}\right)}^{2}$$ (from Eq $$(iii)$$]
    $$\Rightarrow$$ $$36\left(\cfrac{2-x}{2+x}\right)={y}^{2}{\left(\cfrac{4}{2+x}\right)}^{2}$$

    $$\Rightarrow$$ $$36\cfrac{2-x}{2+x}=\cfrac{16{y}^{2}}{{(2+x)}^{2}}$$

    $$\Rightarrow$$ $$36(4-{x}^{2})=16{y}^{2}$$

    $$\Rightarrow$$ $$9(4-{x}^{2})=4{y}^{2}$$

    $$\Rightarrow$$ $$36-9{x}^{2}=4{y}^{2}$$

    $$\Rightarrow$$ $$9{x}^{2}+4{y}^{2}=36$$

    $$\Rightarrow$$ $$\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 9 } =1$$ which represents an ellipse
  • Question 3
    1 / -0
    If the line $$3x+4y=24$$ and $$4x+3y=24$$ intersects the coordinates axes at $$A,B,C$$ and $$D$$, then the equation of the circle passing through these $$4$$ points  is
    Solution
    writing in intercept form, we get 
    $$\dfrac{x}{8}+\dfrac{y}{6}=1$$ and 

    $$\dfrac{x}{6}+\dfrac{y}{8}=1$$
    Hence the point are $$(0,8),(0,6)$$ and $$(6,0)(8,0)$$
    Hence
    The circle cuts the x axis at $$(6,0)(8,0)$$
    Hence x-coordinate of the center will be 
    $$=6+\dfrac{8-6}{2}$$
    $$=7$$
    The circle cuts the x axis at $$(0,6)(0,8)$$
    Hence y-coordinate of the center will be 
    $$=6+\dfrac{8-6}{2}$$
    $$=7$$
    The center of the circle will lie at 
    $$(7,7)=C$$
    Now let the point $$(6,0)$$ be A.
    Hence 
    R=radius
    $$CA$$$$=\sqrt{50}$$
    Hence, the equation of the circle will be 
    $$(x-7)^{2}+(y-7)^{2}=50$$
    $$x^{2}+y^{2}-14x-14y+98=50$$
    Or 
    $$x^{2}+y^{2}-14x-14y+48=0$$
  • Question 4
    1 / -0
    The equation of a straight line drawn through the focus of the parabola $$y^2=-4x$$ at an angle of $$120^o$$ to the $$x$$-axis is.
    Solution
    Parabola, $${ y }^{ 2 }=-4x$$
    Comparing above equation with $${ y }^{ 2 }=-4ax$$
    $$4a=4\Rightarrow a=1$$
    Focus of $${ y }^{ 2 }=-4ax$$ is $$\left( -a,0 \right) $$
    $$\therefore $$ Focus is $$\left( -1,0 \right) $$
    Now, line makes an angle of $${ 120 }^{ 0 }$$ to the $$x$$ - axis
    Therefore, $$m=\tan { \theta  } =\tan { { 120 }^{ 0 } } $$
    $$\Rightarrow m=\tan { \left( { 180 }^{ 0 }-{ 60 }^{ 0 } \right)  } =-\tan { { 60 }^{ 0 } } $$
    $$\Rightarrow m=-\sqrt { 3 } $$
    Equation of line passing through $$\left( -1,0 \right) $$ having slope $$-\sqrt { 3 } $$  is $$:$$
    $$y-0=-\sqrt { 3 } \left( x+1 \right) $$
    $$\Rightarrow \boxed { y+\sqrt { 3 } \left( x+1 \right) =0 } $$
  • Question 5
    1 / -0
    Find the length of latus rectum of the parabola
    $$(a^{2}+b^{2})(x^{2}+y^{2})=(bx+ay-ab)^{2}$$
    Solution
    Alternative Method:  The given equation may be written as
    $$\displaystyle x^{2}+y^{2}=\frac{(bx+ay-ab)^{2}}{(a^{2}+b^{2})}$$
    or  $$\displaystyle \sqrt{x^{2}+y^{2}}=\frac{\left |bx+ay-ab  \right |}{\sqrt{a^{2}+b^{2}}}$$
    or   $$\displaystyle \sqrt{(x-0)^{2}+(y-0)^{2}}=\frac{\left |bx+ay-ab  \right |}{\sqrt{a^{2}+b^{2}}}$$
    which is of the form   $$SP=PM$$
    Since distance from focus S to $$\displaystyle (bx+ay-ab=0)=\frac{1}{2}(4\rho )$$
    or       $$\displaystyle  \frac{1}{2}(4\rho )=\frac{ab}{\sqrt{a^{2}+b^{2}}}$$
    $$\displaystyle 4\rho =\frac{2ab}{\sqrt{a^{2}+b^{2}}}$$
  • Question 6
    1 / -0
    The locus of the point $$(h,k)$$, if the point $$(\sqrt{3h}, \sqrt{3k + 2})$$  lies on the line $$x - y - 1 = 0$$, is a ?
    Solution
    $$(\sqrt{3h}, \sqrt{3k + 1})$$ lines on the line $$x - y - 1 = 0$$
    $$\Rightarrow  (\sqrt{3h})^2 = (\sqrt{3k + 2} + 1)^2$$
    $$\Rightarrow   3h = 3k + 2 + 1 + 2 \sqrt{3 k + 2}$$
    $$\Rightarrow    3^2 (h - k - 1)^2 = 2^2 (\sqrt{3k + 2})^2$$
    $$\Rightarrow    9(h^2 + k^2 + 1 - 2hk - 2h + 2k) = 4(3k + 2)$$
    $$\Rightarrow    9(x^2 + y^2) - 18xy - 18x + 6y + 1 = 0$$
    Now   $$h^2 = ab $$ and $$\Delta \neq 0$$
    Therefore, locus is a parabola.
  • Question 7
    1 / -0
    A circle touches the $$x$$-axis and also touches the circle with centre $$(0, 3)$$ and radius $$2$$. The locus of the centre of the circle is -
    Solution
    Let $$C_1(h, k)$$ be the center of the circle. 

    Circle touches the $$x$$-axis then its radius is $$r_1 = k$$.

    Also circle touches the circle with centre $$C_2(0, 3)$$ and radius $$r_2 = 2$$.

    $$\therefore         |C_1 C_2| = r_1 + r_2$$

    $$\Rightarrow        \sqrt{(h - 0)^2 + (k - 3)^2} = |k+2|$$

    Squaring

    $$h^2 - 10k + 5 = 0$$

    $$\Rightarrow$$ Locus is $$x^2 - 10y + 5 = 0$$, which is parabola.
  • Question 8
    1 / -0
    A point $$(\alpha, \beta)$$ lies on a circle $$x^2+y^2=1$$, then locus of the point $$(3\alpha +2\beta)$$ is a$$/$$an.
    Solution
    Point will be $$(3\alpha ,2\beta )$$ not $$( 3\alpha +2\beta )$$
    Now $$ x^{2}+y^{2}=1 $$
    Radium is $$1$$ unit,hence parametric co - ordinate is 
    $$(\alpha ,\beta )= (1\cos\theta ,1\sin\theta )=(\cos\theta , \sin\theta )$$
    Hence
    Point is $$ (3\cos\theta ,2\sin\theta )$$
    Hence
    $$(x,y)= (3\cos\theta ,2\sin\theta )$$
    $$x=3\cos\theta $$
    $$ \Rightarrow \dfrac{x}{3}\cos\theta$$    ...(i)
    $$ y=2\sin\theta $$
    $$ \Rightarrow \dfrac{y}{2} = \sin \theta$$   ...(ii)
    $$ (i)^{2} + (ii)^{2} $$
    $$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} \cos^{2}\theta + \sin^{2} \theta $$
    $$ \dfrac{x^{2}}{9}+ \dfrac{y^{2}}{4} = 1 $$
    which is equation of ellipse
  • Question 9
    1 / -0
    Find the equation of the circle that passes through the points $$(0,6),(0,0)$$ and $$(8,0)$$
    Solution

    Let the equation of the general form of the required circle be 

    $$x^2+y^2+2gx+2fy+c=0$$................(1)

    According to the problem, the above equation of the circle passes through the points $$(0, 6), (0, 0)$$ and $$(8, 0)$$. Therefore,

    $$36 + 12f + c = 0$$  ………. (2)

    $$ c = 0$$                       ……………. (3)

    $$64 + 16g + c = 0$$  ……………. (4)

    Putting $$c = 0$$ in (2), we obtain $$f = -3$$. Similarly put $$c = 0$$ in (4), we obtain $$g=-4$$

    Substituting the values of $$g, f$$ and $$c$$ in (1), we obtain the equation of the required circle as:

    $$x^2+y^2+2(-4)x+2(-2)y+0=0$$ that is 

    $$x^2+y^2-8x-4y+0=0$$ can be rewritten as

    $$x^2+y^2-8x-4y+16+9=0+16+9$$

    $$(x-4)^2+(y-3)^2=25$$

    Therefore, the equation of circle is $$(x-4)^2+(y-3)^2=25$$.


  • Question 10
    1 / -0
    A circle and a parabola intersect at four points $$(x_1 , y_1), (x_2 , y_2), (x_3 , y_3)$$ and $$(x_4 , y_4)$$. Then $$y_1 + y_2 + y_3 + y_4$$ is equal to
    Solution
    A circle and a parabola intersect at four points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) ;\left( { x }_{ 2 },{ y }_{ 2 } \right) ;\left( { x }_{ 3 },{ y }_{ 3 } \right) ;\left( { x }_{ 4 },{ y }_{ 4 } \right) $$ .
    Then $${ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }$$ is equal to :
    $${ y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }={ 2at }_{ 1 }+{ 2at }_{ 2 }+{ 2at }_{ 3 }+{ 2at }_{ 4 }=0$$ 

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now