3 Marks Question

Question 13 Marks

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
y² - 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
$=4\times\frac{3}{4}$
$=3.$

View full question & answer→

Question 23 Marks

Find the length of the line segment joining the vertex of the parabola y² = 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
$\sqrt{\Big(\frac{\text{4a}}{\tan^2\theta}\Big)+\Big(\frac{\text{4a}}{\tan\theta}\Big)}$
$=\frac{4\text{a}}{\tan^2\theta}\sqrt{1+\tan^2\theta}$
$=4\text{a}\ \cot\theta\text{ cosec }\theta.$

View full question & answer→

Question 33 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 $\text{2y}^2=\text{9x}.$

View full question & answer→

Question 43 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 m₁ 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 $\text{y}^2-\text{4y}-\text{8y}+28=0.$

View full question & answer→

Question 53 Marks

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

Answer

Let (x₁, y₁) 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 $\text{x}^2=-\text{8y}.$

View full question & answer→

Question 63 Marks

At what point of the parabola x² = 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 $(3,1).$

View full question & answer→

MATHS 3 Marks Question

Take a timed test