Solve the Following Question.(5 Marks)

Question 15 Marks

Find the equation of the parabola, if
The focus is at (0, 0) and vertex is at the intersection of the lines x + y = 1 and x - y = 3.

Answer

$\text{x}+\text{y}=1$ and $\text{x}-\text{y}=3$
Intersecting point of above lines is
$\text{(x, y)}=(2,1)......\text{vertex}$
Focus (0, 0)
Vertex is the mid-piont of focus and point on directrix which passes through
$2=\frac{0+\text{x}}{2};-1=\frac{0+\text{y}}{2}$
$\text{(x, y)}=(2, 4)$
Slope of line passing through focus and vertex is $\frac{-1}{2}$
Slope of directrix is 2, as both are perpendicular lines
$\text{y}+2=2(\text{x}-4)$
$\text{2x}-\text{y}=10.....\text{directrix}$
$\text{SP}^2=\text{PM}^2$
$5(\text{x}^2+\text{y}^2)=(\text{2x}-\text{y}-10)^2$
$\text{x}^2+\text{4y}^2-100+\text{4xy}-\text{20y}+\text{40x}=0$
$(\text{x}+\text{2y})^2+20(\text{2x}-\text{y}-5)=0.$

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Question 25 Marks

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
$y^2 = 8x + 8y.$

Answer

The given equation is
$\text{y}^2=\text{8x}+\text{8y}$
$\Rightarrow\text{y}^2-\text{8y}=\text{8x}$
$\Rightarrow\text{y}^2+2\times\text{y}\times4+16=\text{4x}+16$
$\Rightarrow\text{(y}-\text{4})^2=8(\text{x}+2).....\text{(i)}$
Shifting the origin to the point (-2, 4) without rotating the axes and denoting the new coordinates w.r.t these axes by $X$ and $Y,$ we have
$\text{x}=\text{X}-2,\ \text{y}=\text{Y}+4.....\text{(ii)}$
Using these relation equation (i), reduces to $\text{Y}^2=\text{8X}.....\text{(iii)}$
This is of the form $\text{Y}^2=4\text{aX},$ on comparing, we get
$\text{4a} = 8$
$\Rightarrow\text{a}=2$
Now,
Vertex: The coordinates of the vertex w.r.t new axes are $(\text{X}=0,\ \text{Y}=0)$
$\therefore\ \text{x}=0-2,\ \text{y}=0+4$ [Using equation (ii)]
$\Rightarrow\text{x}=0-2,\ \text{y}=4$
$\therefore$ coordinates of the vertex w.r.t old axes are $\Big(-2,4\Big).$
Focus: The coordinate of the focus w.r.t new axes are $(\text{X}=2,\ \text{y}=0)$
$\therefore\text{ x}=-2-2$ and $\text{y}=0-4$ [Using equation (ii)]
$\Rightarrow\text{x}=0,$ and $\text{y}=4$
$\therefore$ coordinate of the focus w.r.t old axes are $(0,4).$
Axis: Equation of the axis of the parabola w.r.t new axes is $\text{y}=0$
$\therefore\text{ y}=0+4 [$Using equation $(ii)]$
$\Rightarrow\text{ y}=4$
$\therefore$ equation of the w.r.t old axes is $\text{y}-4.$
Directrix; Equation of the parabola w.r.t new axes is
$\text{X}=-2$
$\therefore\ \text{x}=-2-2$
$\Rightarrow\text{x}=-4$
$\Rightarrow\text{x}+4=0$
$\therefore$ Equetion of the directrix of the parabola w.r.t old axes is $\text{x}+4=0$
Latus-rectum: The length of the latus-rectum $= 4a$
$=4\times2$
$=8.$

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Question 35 Marks

Find the equation of the parabola, if
The focus is at $(a, 0$) and the vertex is at $(a', 0).$

Answer

In a parabola, vertex is the mid-point of the focus and the point of the intersection of the axis and directrix.
So, let $(x_1, y_1)$ be the co-ordinate of the point of intersection of the axis and directrix.
Then $(a', 0)$ is the mid-point of the line segment joining $(a, 0)$ and $(x_1, y_1).$
$\therefore\frac{\text{x}_1+\text{a}}{2}=\text{a}'$ and $\frac{\text{y}_1-0}{2}=0$
$\Rightarrow\text{x}_1=2\text{a}'-\text{a}$ and $\text{y}_1=0$
Thus, the directrix meets the axis at $(2a'-a, 0)$
So the equation of directrix is $x = 2a' - a$
Let $P(x, y)$ be any point on the parabola.
Then,
$\text{SP}=\text{PM}$
$\Rightarrow\text{SP}^2=\text{PM}^2$
$\Rightarrow\text{(x}-\text{a)}^2+\text{(y}-0)^2=\Big[\frac{\text{x}-\text{2a}'+\text{a}}{\sqrt{1^2}}\Big]^2$
$\Rightarrow\text{x}^2+\text{a}^2-\text{2ax}+\text{y}^2=\text{(x}-\text{2a}'+\text{a})^2$
$\Rightarrow\text{x}^2+\text{a}^2-\text{2ax}+\text{y}^2=\text{x}^2+(-\text{2a}')^2+\text{a}^2+\\2\times\text{x}\times(-\text{2a}')+2\times(-\text{2a}')\times\text{a}+2\times\text{(a)}\times\text{(x)}$
$\Rightarrow\text{x}^2+\text{a}^2-\text{2ax}+\text{y}^2=\text{x}^2+4\text{(a}')^2+\text{a}^2-\text{4xa}'-\text{a}'\text{a}+\text{2ax}$
$\Rightarrow\text{y}^2=\text{x}^2-\text{x}^2+\text{a}^2-\text{a}^2+\text{2ax}+4\text{(a}')^2-\text{4xa}'-4\text{a}'\text{a}+\text{2ax}$
$\Rightarrow\text{y}^2=\text{4ax}-\text{4xa}'+\text{4(a}')\text{a}$
$\Rightarrow\text{y}^2=\text{4ax}-\text{4a}'\text{a}-\text{4xa}'+4\text{(a}')^2$
$=\text{4a(x}-\text{a}')-\text{4a}'\text{(x}-\text{a}')$
$=\text{(4a}-\text{4a}')\text{(x}-\text{a}')$
$=4\text{(a}-\text{a}')\text{(x}-\text{a}')$
$\therefore\text{y}^2=4\text{(a}-\text{a}')\text{(x}-\text{a}')$
$\Rightarrow\text{y}^2=-4\text{(a}'-\text{a})\text{(x}-\text{a}')$
Hence, required equation of parabola is $\text{y}^2=-4\text{(a}'-\text{a})\text{(x}-\text{a}').$

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Question 45 Marks

Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3, 3) and directrix is 3x - 4y = 2. Find also the length of the latus-rectum.

Answer

The axis of the parabola Is a line $\bot$ to the directrix and passing through focus. The equation of a line $\bot$ to $\text{3x}-\text{4y}-2=0$ is
$\text{y}=\frac{-4}{3}+\lambda$ $\begin{bmatrix}\therefore\text{ m}_1\text{m}_2=-1\\\Rightarrow\text{m}_2\frac{-1}{\text{m}_1} \text{ and y =}\text{m}_2\text{x}+\lambda\end{bmatrix}$
$\Rightarrow\ \text{3y}+\text{4x}=3\lambda$
This will pass through focus (3,3) if,
$3\times3+4\times3=3\lambda$
$\Rightarrow\ 9+12=3\lambda$
$\Rightarrow\ 21=3\lambda$
$\Rightarrow\ \lambda=\frac{21}{3}=7$
So, the equation of axis is $\text{3y}+\text{4x}=3\times7=21$
$\Rightarrow\ \text{3y}+\text{4x}=21...\text{(i)}$
And the equation of doreanx is
$\text{3x}-\text{4y}=2....\text{(ii)}$
Mutiplying equation (i) by 4 and equation (ii) by 3, we get
$\text{16x}+\text{12y}=84.....\text{(iii)}$
$\text{9x}-\text{12y}=6....\text{(iv)}$
Adding equation (iii) and (iv), we get
$\text{16x}+\text{9x}=84+6$
$\Rightarrow\ 25\text{x}=90$
$\Rightarrow\ \text{x}=\frac{90}{25}=\frac{18}{5}$
Putting $\text{x}=\frac{18}{5}$ in equation (i), we get
$\text{3y}+4\times\frac{18}{5}=21$
$\Rightarrow\ \text{3y}+\frac{72}{5}=21$
$\Rightarrow\ \text{3y}=21-\frac{72}{5}$
$\Rightarrow\ \text{3y}=\frac{105-72}{5}$
$\Rightarrow\ \text{3y}=\frac{33}{5}$
$\Rightarrow\ \text{y}=\frac{11}{5}$
Hence, the required point of intersection is $\Big(\frac{18}{5},\frac{11}{5}\Big).$

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Question 55 Marks

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:
$4(y - 1)^2 = -7(x - 3).$

Answer

The given equation is
$4(\text{y}-1)^2=-7(\text{x}-3)$
$\Rightarrow(\text{y}-1)^2=\frac{-7}{4}(\text{x}-3)...\text{(i)}$
Shifting the origin to the point (3, 1) without rotating the axes and denoting the new coordinates w.r.t these axes by $X$ and $Y,$ we have,
$\text{x}=\text{X}+3,\ \text{y}=\text{Y}+1....\text{(ii)}$
Using these relation (i), redus to
$\text{Y}^2=\frac{-7}{4}\text{X}......\text{(iii)}$
This is of the form $\text{Y}=-\text{4aX},$ on comparing, we get
$\text{4a}=\frac{7}{4}$
$\Rightarrow\text{a}=\frac{7}{16}$
Now,
Vertex: The coordinates of the vertex w.r.t new axes are $(\text{X}=0,\ \text{Y}=0)$
$\therefore \text{ x}=0+3,\ \text{y}=0+1 [$Using equation $(iii)]$
$\Rightarrow\text{x}=3,\ \text{y}=1$
$\therefore$ coordinate of the vertex w.r.t new axes are $(3, 1).$
Focus: The coordinate of the focus w.r.t new axes are $\Big(\text{x}=-\frac{7}{16},\text{y}=0\Big)$
$\therefore\text{x}-\frac{-7}{16}+3,\ \text{y}=0+1$
$\Rightarrow\text{x}=\frac{41}{16},\ \text{y}=1$
$\therefore$ coordinates of the focus w.r.t old axes are $\Big(\frac{41}{16},1\Big)$
Axis: Equation of the axes of the parabola w.r.t new axes is
$\text{y}=0$
$\Rightarrow\text{y}=0+1$
$\Rightarrow\text{y}=1$
$\therefore$ equation of axis w.r.t old axes is $\text{y}=1$
Directrix: Equation of the directrix of the parabola w.r.t new axes is
$\text{Y}=\frac{7}{16}$
$\therefore\text{x}=\frac{7}{16}+3$
$\Rightarrow\text{x}=\frac{55}{16}$
$\therefore $ Equation of the directrix of the parabola w.r.t oid axes is $\text{x}=\frac{55}{16}.$
Latus-rectum: The length of the latus-rectum $= 4a$
$=4\times\frac{7}{16}$
$=\frac{7}{4}.$

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Question 65 Marks

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

Answer

In the given parabola, $\text{a}=\frac{1}{16}$
Focus $(0,\frac{1}{16})$
vertex $(0, 0)$
Directrix $\text{x,y}=\frac{1}{16}$
Axis, $\text{x} = 0$
$\text{LR}=\frac{1}{4}$
The given equation is
$\text{y}^2-\text{4y}-\text{3x}+1=0$
$\Rightarrow\text{y}^2-\text{4y}=\text{3x}-1$
$\Rightarrow\text{y}^2-\text{4y}+4=\text{3x}-1+4$
$\Rightarrow\text{y}^2-\text{4y}+(2)^2=\text{3x}+3$
$\Rightarrow\text{(y}-\text{2})^2=3\text{(x}+1)....\text{(i)}$
Shifting the origin to the point $(-1, 2)$ without rotating the axis and denote the new coordinates with respect to these axis by $X$ and $Y,$ we have
$\text{x}=\text{X}-1,\text{y}=\text{Y}+2....\text{(ii)}$$$
Using these relations equation (i), reduces to
$\text{y}^2=\text{3X}...\text{(iii)}$
This is of the form $\text{y}^2=\text{4aX}.$
On comparing we get,
$\text{4a}=3$
$\Rightarrow\text{a}=\frac{3}{4}.$
Now,
Vertex: The coordinate of the vertex with respect to new axes are $\text{X}=0,\ \text{Y}=0$
So coordinates of the vertex with respect to old axes are $(-1, 2)$
Focus: The coordinates of the focus with respect to new axes are $\Big(\text{X}=\frac{3}{4},\ \text{Y}=0\Big).$
Putting $\text{X}=\frac{3}{4}$ and $\text{Y}=0$ in equation $(ii),$ we get
$\text{X}=\frac{3}{4}-1$ and $\text{y}=0+ 2$
$\Rightarrow\text{x}=\frac{-1}{4}$ and $\text{y}=2$
$\therefore$ coordinates of the focus of the with respects to old axe are $\Big(\frac{-1}{4},2\Big).$

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Question 75 Marks

Find the equation of the parabola, if
The focus is at $(0, -3)$ and the vertex is at $(-1, -3).$

Answer

In a parabola, vertex is the mid-point of the focus and the point of the intersection of the axis and directrix.
So, let $(x_1, y_1)$ be the co-ordinate of the point of intersection of the axis and directrix.
Then $(-1, -3)$ is the mid-point of the line segment joining $(0, -3)$ and $(x_1, y_1).$
$\therefore\frac{\text{x}_1+0}{2}=-1$ and $\frac{\text{y}_1-3}{2}=-3$
$\Rightarrow\text{x}_1=-2$ and $\text{y}_1=-3$
Thus, the directrix meets the axis at $(-2, -3)$
Let $A$ be the vertex and $S$ be the focus of the required parabola.
Then,
$\text{m}_1=\text{slope of AS}=\frac{-3-(-3)}{0-(-1)}=0$
$\therefore$ slope of the directrix $=\frac{-1}{0}=\infty$
Thus, the directrix passes through (-2, -3) and has slope $\infty,$ so its equation is
$\text{y}-(-3)=\infty(x-(-2))$
$\frac{\text{y}+3}{\infty}=\text{x}+2$
$\Rightarrow\text{x}+2$
Let $P(x, y)$ be a point on parabola.
Then, $PS =$ Distance of $P$ from the directrix.
$\sqrt{\text{(x}-2)^2+\text{(y}+3)^2}=\Big|\frac{\text{x}+2}{\sqrt{1^2}}\Big|$
$\Rightarrow\text{x}^2+\text{(y}+3)^2=\text{(x}+2)^2$
$\Rightarrow\text{x}^2+\text{y}^2+9+\text{6y}=\text{x}^2+4+\text{4x}$
$\Rightarrow\text{y}^2-\text{4x}+\text{6y}+9-4=0$
$\Rightarrow\text{y}^2-\text{4x}+\text{6y}+5=0.$

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Question 85 Marks

Find the equation of the parabola whose focus is the point (2, 3) and directrix is the line x - 4y + 3 = 0. Also, find the length of its latus-rectum.

Answer

Let P(X, Y) be any point on the parabola whose focus is S (2, 3) and the directrix x - 4y + 3 = 0. Draw PM perpendicular from P(X, Y) on the directrix x - 4y + 3 = 0.
Then by definition
$\text{SP}=\text{PM}$
$\Rightarrow\text{SP}^2=\text{PM}^2$
$\Rightarrow\text{(x}-2)^2+\text{(y}-3)^2=\Bigg(\frac{\text{x}-\text{4y}+3}{\sqrt{1^2+(-4)^2}}\Bigg)^2$
$\Rightarrow\text{x}^2+4-\text{4x}+\text{y}^2+9-\text{6y}=\frac{\text{(x}-\text{4y}+3)^2}{(\sqrt{17})^2}$
$\Rightarrow\text{x}^2+\text{y}^2-\text{4x}-\text{6y}+4+9=\frac{(\text{x}-\text{4y}+3)^2}{17}$
$\Rightarrow\text{17(x}^2+\text{y}^2-\text{4x}-\text{6y}+13)=\text{(x}-\text{4y}+3)^2$
$\Rightarrow\text{17x}^2+\text{17y}^2-\text{68x}-\text{102y}+122=\text{x}^2+(-\text{4y})^2\\+3^2+2\times\text{x}\times(-\text{4y)}+2\times(-\text{4y)}\times3+2\times3\times\text{x}$
$\Rightarrow\text{17x}^2+\text{17y}^2-\text{68x}-\text{102y}+221=\text{x}^2+\text{16y}^2+9-\text{8xy}-\text{24y}+\text{6x}$
$\Rightarrow\text{17x}^2-\text{x}^2+\text{17y}^2-\text{16}^2+\text{8xy}-\text{68x}-\text{6x}-\text{102y}+\text{24y}+221-9=0$
$\Rightarrow\text{16x}^2+\text{y}^2+\text{8xy}-\text{74x}-\text{78y}+212=0$
This is the questions of the required parabola.
Latus Rectum = Legnth of perprndicular from focus (2, 3) on directrix x - 4y + 3 = 0
$=2\Big|\frac{2-12+3}{\sqrt{1+16}}\Big|$
$=2\Big|\frac{-7}{\sqrt{17}}\Big|$
$=\Big|\frac{14}{\sqrt{17}}\Big|.$

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Maths Solve the Following Question.(5 Marks)

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