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Conic Sections Test - 18

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Conic Sections Test - 18
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  • Question 1
    1 / -0
    Two vertices of an equilateral triangle are $$(-1, 0)$$ and $$(1, 0)$$, and its third vertex lies above the $$x$$-axis. The equation of the circumcircle of the triangle is
    Solution
    Points $$A(-1,0)$$ and $$B(1,0)$$ lie on $$x$$-axis.

    Mid-point of $$AB$$ is $$(0,0)$$.

    Clearly, point $$C$$ lies on $$y$$-axis. Since, it lies above $$y$$-axis. 
    So, let the coordinates of point $$C$$ be $$(0, k)$$.

    Coordinates of centroid $$G$$ is $$\displaystyle \left (0,\dfrac{k}{3}\right)$$.
    We also know that circumcenter, centroid and incenter of the equilateral triangle coincides. 

    Hence, the coordinates of circumcenter is $$\displaystyle \left (0,\dfrac{k}{3}\right)$$.

    Since, $$ABC$$ is equilateral triangle,

    $$AB=AC$$

    $$\Rightarrow (AB)^2=(AC)^{2}$$

    $$\Rightarrow k=\sqrt{3}$$

    So, the coordinates of circumcenter is $$\displaystyle \left (0,\dfrac{1}{\sqrt{3}}\right)$$.

    Radius $$CG=\displaystyle \dfrac{2}{\sqrt{3}}$$

    Equation of circumcircle is

    $$(x-0)^{2}+\left (y-\dfrac{1}{\sqrt{3}}\right)^{2}=\displaystyle \dfrac{4}{3}$$

    $$\Rightarrow x^{2}+y^2-\dfrac{2}{\sqrt{3}}y-1=0$$

    $$\Rightarrow \displaystyle \sqrt{3}\left ( x^{2}+y^{2} \right )-2y-\sqrt{3}=0$$

  • Question 2
    1 / -0
    The length of the latus rectum of the parabola whose focus is $$\left ( 3,3 \right )$$ and directrix is  $$3x-4y-2=0$$ is
    Solution
    Length of the latus rectum $$(l)=2($$perpendicular distance from focus to directix$$)$$
    $$l=2\left| \dfrac { 3\left( 3 \right) -4\left( 3 \right) -2 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } }  }  \right|=2 $$
  • Question 3
    1 / -0
    The triangle PQR is inscribed in the circle $$\displaystyle x^{2}+y^{2}= 25.$$ If $$Q$$ and $$R$$ have coordinates $$\displaystyle \left ( 3, 4 \right )$$ and $$\displaystyle \left ( -4, 3 \right ),$$ respectively, then $$\displaystyle \angle QPR$$ is equal to
    Solution
    Now, length of chord $$QR=\sqrt{(-4-3)^2+(3-4)^2}=5\sqrt{2}$$

    Also, $$OR=OQ=5$$ (radii of the circle)

    $$\therefore \angle ROQ=90^{\circ}$$

    Using the theorem that arc subtended by a chord is twice the angle subtended by the chord at the arc of the circle,

     we have $$\angle QPR=45^{\circ}=\dfrac{\pi}{4}$$

  • Question 4
    1 / -0
    The lines $$\displaystyle 2x - 3y = 5$$ and $$\displaystyle 3x - 4y = 7$$ intersect at the center of the circle whose area is $$154$$ sq. units, then equation of circle is
    Solution
    Since, the lines $$\displaystyle 2x - 3y = 5 \dots (1)$$ and $$\displaystyle 3x - 4y = 7 \dots (2)$$ passes through the center of the circle.

    Multiplying $$(1)$$ by 4 and $$(2)$$ by 3. Further subtracting we get,

    $$-1x=-1 \Rightarrow x=1$$

    Substituting $$x=1$$ in $$(1)$$. We get,

    $$y=-1$$

    So, point of intersection of these two lines is $$(1,-1)$$

    So, center is at $$(1,-1)$$

    Area of circle $$=\pi r^2$$

    $$154=\dfrac{22}{7} r^2$$

    $$r^2=49$$

    Equation of circle is $$(x-1)^2+(y+1)^2=r^2$$

    $$(x-1)^2+(y+1)^2=49$$

    $$x^2-2x+1+y^2-2y+1=49$$

    $$\Rightarrow x^2+y^2-2x+2y=47$$
  • Question 5
    1 / -0
    The equation of circle with origin as a centre and passing through equilateral triangle whose median is of length $$3a$$ is
    Solution
    Given, median of the equilateral triangle is $$3a$$ say $$LD $$
    In $$\displaystyle \Delta LMD$$, we have
    $$(LM)^{2}=(LD)^{2}+(MD)^{2}$$
    $$\Rightarrow \displaystyle (LM)^{2}=9a^{2}+\left(\frac{LM}{2}\right)^{2} $$ 
    $$\displaystyle\Rightarrow \frac{3}{4}(LM)^{2}=9a^{2}$$
    $$\Rightarrow (LM)^{2}=12a^{2}$$ 
    Again in triangle $$\displaystyle OMD$$,
    $$(OM)^{2}=(OD)^{2}+(MD)^{2} $$
    $$\Rightarrow \displaystyle R^{2}=(3a-R)^{2}+ \left(\frac{LM}{2}\right)^{2}$$
    $$\displaystyle\Rightarrow R^{2} =9a^{2}+R^{2}-6aR+3a^{2} $$ 
    $$\displaystyle \Rightarrow 6aR=12a^{2}$$
    $$\Rightarrow  R=2a$$
    So, equation of circle is
    $$\displaystyle (x-0)^{2}+(y-0)^{2}=(2a)^{2}$$
    $$ \Rightarrow x^{2}+y^{2}=4a^{2}$$

  • Question 6
    1 / -0
    The equation of a diameter of a circle is $$x+y=1$$ and the greatest distance of any point of the circle from the diameter is $$\dfrac{1}{\sqrt{2}}$$ .Then, a possible  equation of the circle can be
    Solution
    Given diameter of circle is $$x+y=1$$
    Diameter crosses the x-axis at $$(1,0)$$ and y-axis at $$(0,1)$$
    Distance between these two points i.e. $$d=\sqrt{2}$$
    $$\Rightarrow \displaystyle r=\frac{\sqrt{2}}{2}$$
    Center of the circle is $$\displaystyle (\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } )$$
    Equation of circle is 
    $$\displaystyle { (x-\frac { 1 }{ 2 } ) }^{ 2 }+{ (y-\frac { 1 }{ 2 } ) }^{ 2 }=\frac { 1 }{ 2 } $$
    $$x^{2}+y^{2}-x-y=0$$
  • Question 7
    1 / -0
    Find the  Lactus Rectum of  $$\displaystyle 9y^{2}-4x^{2}=36$$ 
    Solution
    $$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$$ 
    Here the coefficient of $$\displaystyle y^{2}$$ is + ive and that of $$\displaystyle x^{2}$$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.
    $$\displaystyle a^{2}=4, b^{2}=9.$$
    $$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$$
    or $$\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$$ 
    Foci lie on y-axis $$\displaystyle \left ( 0, \pm ae \right )$$ i.e $$\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$$ 
    $$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$$
  • Question 8
    1 / -0
    If the centroid of an equilateral triangle is $$(1, 1)$$ and its one vertex is $$(-1, 2)$$ then the equation of its circumcircle is
    Solution
    Given centroid of an equilateral triangle is $$G(1,1)$$.
    We know that in an equilateral triangle, centroid, circumcenter and incenter are at the same point.
    So, the circumcenter is at $$G(1,1)$$.
    Given one vertex of equilateral triangle at $$A(-1,2)$$
    So, circumradius $$=AG=\sqrt { 5 } $$
    So, equation of circumcircle is
    $$(x-1)^2+(y-1)^2=5$$
    $$\Rightarrow x^{2}+y^{2}-2x-2y-3=0$$
  • Question 9
    1 / -0
    The equation $$ \displaystyle 3x^{2}-2xy+y^{2}=0 $$ represents:
    Solution
    Given expression,$$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$$ 
    As Coefficient of $$\displaystyle xy$$ is not zero,It will not be a circle and hyperbola.
    Let $$\displaystyle \frac { y }{ x } =m$$
    We get $$\displaystyle { m }^{ 2 }-2m+3=0$$ will not have any real solutions as discriminant is less than zero.
    $$\displaystyle \therefore $$ They will not be pair of lines too.
  • Question 10
    1 / -0
    The length of the latus rectum of the parabola $$x=ay^2+by+c$$ is 
    Solution
    Given,

    $$x=ay^2+by+c$$

    $$ay^2+by=x-c$$

    $$a\left ( y+\dfrac{b}{a} \right )^2=x+\dfrac{ab^2}{4}-c$$

    $$\left ( y+\dfrac{b}{a} \right )^2=\dfrac{1}{a}\left ( x+\dfrac{ab^2}{4}-c \right )$$

    Length of latus rectum $$=\dfrac{1}{a}$$
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