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This is one of our many quizzes on Parabola. Click on this to access more such competitive quizzes on Conic Sections.

Parabola Quiz 2

mathematics

 Find the equation of the parabola whose vertex is at $(0,4)$ and focus is the origin.

 To find the equation of parabola, find the general term and then put the values.

We have, Vertex $(0,4)$ and Focus $(0,0)$ $ \\ $ General equation would be, $ \\ \Rightarrow (y - a) =  - 4a{x^2} \\ $ Putting the values in the general equation, $ \\ \Rightarrow (y - 2) =  - 4 \times 2 \times {x^2} \\ \Rightarrow (y - 2) =  - 8{x^2}$

 If the ends of the latus rectum of a Parabola are $(8,4)$ and $( - 8,4)$ find the equation of the Parabola.

 Find focus to form an equation and focus is the midpoint Latus rectum.

Latus rectums are in $1^{st}$ and $2^{nd}$ Quadrant, which portrays Latus Rectum as horizontal. $ \\ $ The general equation would be, $ \\ \Rightarrow {x^2} = 4ay \\ $  The mid-point of Latus rectum is the focus, $ \\ \Rightarrow Focus = (0,a) = \left( {\dfrac{{ - 8 + 8}}{2},\dfrac{{4 + 4}}{2}} \right) \\ \Rightarrow (0,4) \\ \Rightarrow a = 4$  Hence, The equation would be, $ \\ \Rightarrow {x^2} = 4 \times 4 \times y \\ \Rightarrow {x^2} = 16y$

 Form the equation of parabola which has : Focus $(0, - 4)$; Directrix $y = 4$ 

 To form the equation of any parabola,  find the general term.

Focus $(0, - 4)$and Directrix $y = 4$ $ \\ $ Since the coordinates of focus are in the form $(0, - a)$  $ \\ $ The general form of the equation would be  $ \\ \Rightarrow {x^2} =  - 4ay \\ $  Also, $ \Rightarrow a = 4$  $ \\ $ Hence, The equation of parabola would be $ \\ \Rightarrow {x^2} =  - 4 \times 4 \times y \\ \Rightarrow {x^2} =  - 16y$

 Find the directrix of the equation: ${x^2} - 12x - 24y + 48 = 0$ 

 To find the equation of directrix just write the general formula and find the missing terms.

$ \Rightarrow {x^2} - 12x - 12y + 48 = 0 \\ $ General form  is ${(x - h)^2} =  \pm 4a(y - k) \\ $ Rewrite the equation in general form $ \\ \Rightarrow {x^2} - 12x + 48 = 12y \\ \Rightarrow ({x^2} - 2 \times 6 \times x + {6^2}) + 12 = 12y \\ \Rightarrow {x^2} - 2 \times 6 \times x + {6^2} = 12y - 12 \\ \Rightarrow {(x - 6)^2} = 12(y - 1) \\ \Rightarrow {(x - 6)^2} = 4 \times 3 \times (y - 1) \\ \Rightarrow h = 6\;\;and\;\;k = 1 \\ \Rightarrow a = 3 \\ $  The equation of directrix is given by $y = k - a$  $ \\ \Rightarrow y = 1 - 3 \\ \Rightarrow y =  - 2$ 

 Find the area of the figure which is formed by joining the vertex of the parabola and latus rectum of the parabola ${y^2} = 40x$ .

 To find the area of the triangle, draw a clear diagram to make the question easy.

Given $ \Rightarrow {y^2} = 40x$ $ \\ $ then the coordinates would be $(0,a)$ $ \\ $ To find $a$, compare it with the general equation ${y^2} = 4ax$ $ \\ $ Rewriting the equation as, $ \\ \Rightarrow {y^2} = 4 \times 10 \times x \\ \therefore a = 10 \\ $ The focus point is $(0,10)$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ea7b8802-5c68-400b-bf74-f709371437e78282056477426325855.png"/> $ \\ $  The ends of its latus rectum clearly are : $ \\ \Rightarrow A( - 2a,a)\;\;and\;\;B(2a,a) \\ \Rightarrow A( - 20,10)\;\;and\;\;B(20,10) \\ $  The figure obtained is a triangle, So area of the triangle would be:  $ \\ \Rightarrow \dfrac{1}{2} \times b \times h \\ \Rightarrow \dfrac{1}{2} \times 40 \times 10  \Rightarrow 40units$ 