3 Marks Question

Question 13 Marks

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
$y^2 - 4y - 3x + 1 = 0.$

Answer

Axis: Equation of the parabola w.r.t new axes is $\text{Y} = 0$
$\therefore\text{y}=0+2$
$\Rightarrow \text{y}=2$
$\therefore$ equation of axis w.r.t old axes is $\text{y}= 2$
Directrix: Equation of the directrix of the parabola w.r.t new axes is $\text{X}=\frac{-3}{4}$
$\therefore\text{x}=\frac{-3}{4}-1$
$\Rightarrow \text{x}=\frac{-7}{4}$
$\therefore$ Equation of the directrix of the parabola w.r.t old axes is $\text{X}=\frac{-7}{4}$
Latus-rectum: The length of the latus-rectum $= 4a$
$=4\times\frac{3}{4}$
$=3.$

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Question 23 Marks

Find the length of the line segment joining the vertex of the parabola $y^2 = 4ax$ and a point on the parabola where the line-segment makes an angle $\theta$ to the x-axis.

Answer

$\text{y}=\text{x}\tan\theta$
$\text{y}^2=\text{4ax}$
Intersection point of both the curves are $\Big(\frac{\text{4a}}{\tan^2\theta},\frac{\text{4a}}{\tan\theta}\Big)$
So Distance from origin to the above point is
$\sqrt{\Big(\frac{\text{4a}}{\tan^2\theta}\Big)+\Big(\frac{\text{4a}}{\tan\theta}\Big)}$
$=\frac{4\text{a}}{\tan^2\theta}\sqrt{1+\tan^2\theta}$
$=4\text{a}\ \cot\theta\text{ cosec }\theta.$

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Question 33 Marks

At what point of the parabola $x^2 = 9y$ is the abscissa three times that of ordinate$?$

Answer

Let the ordinates of the required point is $y.$
$\therefore\text{ abscissa}=3\text{y}$
$\therefore$ The coordinates of the points are $\text{(3y, y)}.$
These points lines on the parabola $\text{x}^2=\text{9y}.$
$\therefore\ \text{(3y)}^2=\text{9y}$
$\Rightarrow\ \text{9y}^2=\text{9y}$
$\Rightarrow\ \text{9y}^2-\text{9y}=0$
$\Rightarrow\ \text{9y}\text{(y}-1)=0$
$\Rightarrow\ \text{y}-1=0$ $\big[\therefore\text{y}\not=0\big]$
$\Rightarrow\ \text{y}=1$
$\therefore\text{abcissa}=3\times\text{y}=3$
Hence, the required point Is $(3,1).$

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Question 43 Marks

Find the equation of a parabola with vertex at the origin, the axis along x-axis and passing through (2, 3).

Answer

Let the equation of parabola be
$\text{y}^2=\text{4ax}.....\text{(i)}$ $[\therefore$ axis along x-axis $]$
If passes through $(2,3)$
$\therefore\ (3)^2=4\times\text{a}\times2$
$\Rightarrow\ 9=\text{8a}$
$\Rightarrow\ \text{a}=\frac{9}{8}$
Putting the value of a in equation (i), we get
$\text{y}^2=4\times\frac{9}{8}\times\text{x}$
$\Rightarrow\ \text{y}^2=\frac{9}{2}\times\text{x}$
$\Rightarrow\text{2y}^2=\text{9x}$
Hence, the required equation of parabola is $\text{2y}^2=\text{9x}.$

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Question 53 Marks

Find the equation of the parabola whose focus is $(5, 2)$ and having vertex at $( 3, 2).$

Answer

In a parabola, vertex is the mid point of the focus and the point of intersection of the axis and directrix. So let $(x, y)$ be the coordinates of the point of intersection of the axis and directrix.
Then $(3, 2)$ is lhemid poim of ihelinesegmentjoining $(5, 2)$ and $(x, y).$
$\frac{\text{x}_1+5}{2}=3$ and $\frac{\text{y}_1+2}{2}=2$
$\text{x}_1+5=6$ and $\text{y}_1+2=4$
$\text{x}_1=6$ and $\text{y}_1=2$
The directrix meets the axis at $(1, 2)$
Let A be the vertex and S be the focus of the required parabola
Then
$\text{m}_1\text{ slope of AS}=\frac{2-2}{5-3}=0$
Let $m_1$ be the slope of the dirtclrix
Then
$\text{m}_2=\infty$ $[\therefore$ Directrix is perpendicular to the axis $]$
Thus the directrix passes througb (1, 2) and the slope $\infty,$ so is equation is
$\text{y}-2=\infty(\text{x}-1)$
$\frac{\text{y}-2}{\infty}=\text{(x}-1)$
$\text{x}-1=0$
Let $P(x, y)$ be a point on the parabola
Then $PS =$ distance of $P$ from the dirtctrix
$\sqrt{\text{(x}-5)^2+\text{(y}-2)^2}=\Big|\frac{\text{x}-1}{\sqrt{1^2}}\Big|$
$\text{(x}-5)^2+\text{(y}-2)^2=\text{(x}-1)^2$
$\text{x}^2+25-\text{10x}+\text{y}^2+4-\text{}\text{4y}=\text{x}^2+1-\text{2x}$
$\text{y}^2-\text{4y}-\text{8y}+28=0$
Hence the required equation of the parabola is $\text{y}^2-\text{4y}-\text{8y}+28=0.$

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Question 63 Marks

Find the equation of a parabola with vertex at the origin and the directrix, $y = 2.$

Answer

Let $(x_1, y_1)$ be the coordinates of the point intersection of the axis and the directrix.
$\therefore\ (\text{x}_1,\text{ y}_1)=(0,\ 2)$
we know that, the vertex is the mid-point of the fine segment joining $(0, 2)$ and focus $(\text{x}_1,\ \text{y}_1)$
$\therefore\ \frac{\text{x}_2+0}{2}=0$ and $\frac{\text{y}_2+2}{2}=0$
$\therefore\ \text{x}_2=0$ and $\text{y}_2=-2$
$\therefore$ The coordinates of focus is $(0,-2)$
By the definition of parabola
$\text{PS}=\text{PM}$
$\Rightarrow\ \text{PS}^2=\text{PM}^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y}+2)^2=\Big[\frac{\text{y}-2}{\sqrt1}\Big]^2$
$\Rightarrow\ \text{x}^2+\text{y}^2+4+\text{4y}=(\text{y}-2)^2$
$\Rightarrow\ \text{x}^2+\text{y}^2+4+\text{4y}=\text{y}^2+4-4\text{y}$
$\Rightarrow\ \text{x}^2=-\text{4y}-\text{4y}$
$\Rightarrow\ \text{x}^2=-\text{8y}$
Hence, The required equation of parabola is $\text{x}^2=-\text{8y}.$

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