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This is one of our many quizzes on Circles and ellipses. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Circles and ellipses Quiz 2

mathematics

 Rasha was studying about ellipses and has her concepts mixed up. Help her by telling which of the fact about ellipses is wrong

 Match all the options with the properties of the conic section to get the correct answer.

  If a plane is parallel to the axis of cone, cut the cone then the cross-section gives hyperbola $ \\ $ If the plane is parallel to base it gives circle $ \\ $ If the plane is inclined with an angle more than the external angle of cone it gives parabola $ \\ $ If the plane is inclined and cut every generators then it forms an ellipse $ \\ $ Other options describe the properties of ellipse and are correct

 An Engineer wants to design a component which is circular in shape with its centre present at $(1,1)$ and one point of its circumference at $(-1,-3)$. If he wishes to find the opposite point at the circumference from the given point it will be?

 The diameter of the circle will be 2 times the radius, use the midpoint formula to find the other coordinate.

  $  \text{Using midpoint formula of circle} \\  \text{x=}\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \\  \Rightarrow \text{1=}\left( \dfrac{-1+{{x}_{2}}}{2} \right) \\  \Rightarrow {{x}_{2}}\text{=3} \\  \text{Similarly, for y coordinate} \\  \text{y=}\left( \dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\  \Rightarrow \text{1=}\left( \dfrac{-3+{{y}_{2}}}{2} \right) \\  \Rightarrow {{y}_{2}}\text{=5} \\  \text{Thus the point is (3,5)} \\ $

 The locus of a particle is such that the space between its directrices is three times compared to the distance between the foci. What can you say about the eccentricity of the locus made by the particle?

 Form an equation from the given data in the question using the formula, The distances between Directrices is given as $D=\dfrac{2a}{e}$  and distance between the foci is $2ae$. 

The locus follows the curve of an ellipse, The distances between Directrices is given as $ \text{D=}\dfrac{2a}{e} \\ $  Where  $a$  s the length of semi major axis $e$ is the eccentricity $ \\ $ Distance between the foci is $2ae$ $ \\ $ Given distance between Directrices$=2\times (\text{Distance between the foci})$ $\text{Therefore.} \dfrac{2a}{e}=3\times 2ae \\  \text{or }{{e}^{2}}=\dfrac{1}{3} \\  \Rightarrow e=\dfrac{1}{\sqrt{3}}$

 Raj tries to rotate a circle given by the equation ${{x}^{2}}+{{y}^{2}}-4x+2y-20=0$  with it’s axis as diameter to make a solid sphere, find the volume of the sphere.

 The equation represents a circle, convert it into the general form of the circle and compare to find the radius. Find the volume using this radius.

 $  \text{The equation of the curve is given as} \\  {{x}^{2}}+{{y}^{2}}-4x+2y-20=0 \\  \text{This represents the equation of a circle} \\  \text{By completing the square} \\  {{x}^{2}}+{{y}^{2}}-2x-4y+{{(1)}^{2}}+{{(2)}^{2}}=20+{{(1)}^{2}}+{{(2)}^{2}} \\  {{(x-2)}^{2}}+{{(y+1)}^{2}}=25={{5}^{2}} \\  \text{we know that the general equation of circle is } \\  {{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}} \\  \text{Comparing the equation with its general form,} \\  \text{Centre of the curve}\equiv \text{(2,-1)} \\  \text{Radius of the circle will be 5 units} \\  \text{Therefore the radius of the sphere made is 5 units} \\  \text{Volume of sphere=}\dfrac{4}{3}\pi {{r}^{3}} \\  \text{Volume of sphere=}\dfrac{4}{3}\pi {{5}^{3}}=523.6unit{{s}^{3}} \\ $

 Suppose you want to connect the circumference of the circle ${{x}^{2}}+{{y}^{2}}+4x-6y=12$  and the point $(3,8)$. What would be the shortest length of the wire to connect them? 

 The shortest distance is shown by the perpendicular line drawn between the point and circle.

 The equation of the circle is given as $ \\  {{x}^{2}}+{{y}^{2}}+4x-6y=12 \\ $ By completing the square $ \\  \Rightarrow {{x}^{2}}+{{y}^{2}}+2x+4y+{{(2)}^{2}}+{{(3)}^{2}}=12+{{(2)}^{2}}+{{(3)}^{2}} \\  \Rightarrow {{(x+2)}^{2}}+{{(y-3)}^{2}}=25={{5}^{2}} \\ $ We know that the general equation of circle is $ \\  {{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}} \\ $ The centre of circle lies at $\equiv \text{(-2,3)}$ $ \\ $ The shortest distance is shown by the perpendicular line drawn between the point and circle $ \\ $ The distance between two points is given as $ \\  \text{d=}\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ \text{d=}\sqrt{{{(3-(-2))}^{2}}+{{(8-3)}^{2}}} \\  \text{d=7}\text{.1 units} \\ $ Let the distance between the circumference and point be $x$ $ \\ \text{x=d-radius} \\  \text{x=7}\text{.1-5} \\  \text{x=2}\text{.1 units} \\ $ 