Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 25 - Parabola

Answer:

(i) Let P (x, y) be any point on the parabola whose focus is S (3, 0) and the directrix is 3x + 4y = 1.
Draw PM perpendicular to 3x + 4y = 1.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-0\right)}^2={\left(\frac{3x+4y-1}{\sqrt{9+16}}\right)}^2 \\ \Rightarrow {\left(x-3\right)}^2+y^2={\left(\frac{3x+4y-1}{5}\right)}^2 \\ \Rightarrow 25\left\{{\left(x-3\right)}^2+y^2\right\}={\left(3x+4y-1\right)}^2 \\ \Rightarrow \left(25x^2-150x+25y^2+225\right)=9x^2+16y^2+1+24xy-8y-6x \\ \Rightarrow 16x^2+9y^2-24xy-144x+8y+224=0 \end{gathered}$$

(ii) Let P (x, y) be any point on the parabola whose focus is S (1, 1) and the directrix is x + y + 1 = 0.
Draw PM perpendicular to x + y + 1 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2={\left|\frac{x+y+1}{\sqrt{1+1}}\right|}^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2={\left(\frac{x+y+1}{\sqrt{2}}\right)}^2 \\ \Rightarrow 2\left(x^2+1-2x+y^2+1-2y\right)=x^2+y^2+1+2xy+2y+2x \\ \Rightarrow \left(2x^2+2-4x+2y^2+2-4y\right)=x^2+y^2+1+2xy+2y+2x \\ \Rightarrow x^2+y^2-2xy-6x-6y+3=0 \end{gathered}$$

(iii) Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is 2x − y − 1 = 0.
Draw PM perpendicular to 2x − y − 1 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left|\frac{2x-y-1}{\sqrt{4+1}}\right|}^2 \\ \Rightarrow x^2+y^2={\left(\frac{2x-y-1}{\sqrt{5}}\right)}^2 \\ \Rightarrow 5x^2+5y^2=4x^2+y^2+1-4xy+2y-4x \\ \Rightarrow x^2+4y^2+4xy-2y+4x-1=0 \end{gathered}$$

(iv) Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.
Draw PM perpendicular to x − 4y + 3 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left|\frac{x-4y+3}{\sqrt{1+16}}\right|}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^2 \\ \Rightarrow 17\left(x^2+4-4x+y^2-6y+9\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow \left(17x^2-68x-102y+17y^2+13\times 17\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow 16x^2+y^2+8xy-74x-78y+212=0 \end{gathered}$$

Answer:

Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.

Draw PM perpendicular to x − 4y + 3=0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left(\frac{x-4y+3}{\sqrt{1+16}}\right)}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^2 \\ \Rightarrow 17\left(x^2+4-4x+y^2-6y+9\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow \left(17x^2-68x+17y^2-102y+13\times 17\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow 16x^2+y^2+8xy-74x-78y+212=0 \end{gathered}$$

Length of the latus rectum = 2(Length of the perpendicular from the focus on the directrix)
                                           = 2(Length of the perpendicular from (2, 3) on the directrix)
                                           = $2\left|\frac{2-12+3}{\sqrt{16+1}}\right|=2\left|\frac{-7}{\sqrt{17}}\right|=2\left(\frac{7}{\sqrt{17}}\right)=\frac{14}{\sqrt{17}}$ units

Answer:

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (x₁, y₁) be the coordinates of the point of intersection of the axis and directrix.

(i) It is given that the vertex and the focus of a parabola are (−2, 2) and (−6, −6), respectively.

∴ Slope of the axis of the parabola = $\frac{-6-2}{-6+2}=\frac{-8}{-4}=2$

Slope of the directrix = $\frac{-1}{2}$

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r-6}{2}=-2, \frac{s-6}{2}=2 \\ \Rightarrow r=2, s=10 \end{gathered}$$

∴ Required equation of the directrix:

    $y-10=\frac{-1}{2}\left(x-2\right)$
⇒ $2y+x-22=0$.

Now, let P (x, y) be any point on the parabola whose focus is S (−6, −6), and the directrix is $2y+x-22=0$.

Draw PM perpendicular to $2x+y+22=0$.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x+6\right)}^2+{\left(y+6\right)}^2={\left(\frac{2y+x-22}{\sqrt{5}}\right)}^2 \\ \Rightarrow 5\left(x^2+12x+36+y^2+12y+36\right)=4y^2+x^2+484+4xy-88y-44x \\ \Rightarrow 4x^2+y^2-4xy+104x+148y-124=0 \\ \Rightarrow {\left(2x-y\right)}^2-4\left(26x+37y-31\right)=0 \end{gathered}$$

(ii) It is given that the vertex and the focus of a parabola are (0, 0) and (0, −3), respectively.

Thus, the slope of the axis of the parabola cannot be defined.

 Slope of the directrix = 0

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+0}{2}=0,\frac{s-3}{2}=0 \\ \Rightarrow r=0, s=3 \end{gathered}$$

∴ Required equation of directrix:
$y=3$

Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $y=3$.

Draw PM perpendicular to $y=3$.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y+3\right)}^2={\left(\frac{y-3}{\sqrt{1}}\right)}^2 \\ \Rightarrow x^2+y^2+6y+9=y^2-6y+9 \\ \Rightarrow x^2=-12y \end{gathered}$$


(iii) It is given that the vertex and the focus of a parabola are (−1, −3) and (0, −3), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+0}{2}=-1,\frac{s-3}{2}=-3 \\ \Rightarrow r=-2, s=-3 \end{gathered}$$

∴ Required equation of the directrix:
$x+2=0$

Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $x+2=0$.

Draw PM perpendicular to $x+2=0$.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y+3\right)}^2={\left|\frac{x+2}{\sqrt{1}}\right|}^2 \\ \Rightarrow x^2+y^2+6y+9=x^2+4x+4 \\ \Rightarrow y^2+6y-4x+5=0 \end{gathered}$$


(iv) It is given that the vertex and the focus of a parabola are (a', 0) and (a, 0), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+a}{2}=a\text{'}, \frac{s+0}{2}=0 \\ \Rightarrow r=2a\text{'}-a, s=0 \end{gathered}$$

∴ Required equation of the directrix is $x-2a\text{'}+a=0$.

Let P (x, y) be any point on the parabola whose focus is S (a, 0), and the directrix is $x-2a\text{'}+a=0$.

Draw PM perpendicular to $x-2a\text{'}+a=0$.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-a\right)}^2+{\left(y-0\right)}^2={\left(\frac{x-2a\text{'}+a}{\sqrt{1}}\right)}^2 \\ \Rightarrow y^2={\left(x-2a\text{'}+a\right)}^2-{\left(x-a\right)}^2 \\ \Rightarrow y^2=x^2+4a{\text{'}}^2+a^2-4a\text{'}x-4aa\text{'}+2ax-x^2-a^2+2ax \\ \Rightarrow y^2=4a{\text{'}}^2-4a\text{'}x-4aa\text{'}+4ax \\ \Rightarrow y^2=-4\left(a\text{'}-a\right)\left(x-a\text{'}\right) \end{gathered}$$


(v) The point of intersection of $x+y=1 \mathrm{and} x-y=3$ is (2, −1).

Thus, the vertex and the focus of the parabola are (2, −1) and (0, 0), respectively.

∴ Slope of the axis of the parabola = $\frac{0+1}{0-2}=\frac{-1}{2}$

The slope of the directrix is 2.

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+0}{2}=2,\frac{s+0}{2}=-1 \\ \Rightarrow r=4, s=-2 \end{gathered}$$

The required equation of the directrix is $y+2=2\left(x-4\right)$, which can be rewritten as $y-2x+10=0$.

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and directrix is $y-2x+10=0$.

Draw PM perpendicular to $y-2x+10=0$.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(\frac{y-2x+10}{\sqrt{5}}\right)}^2 \\ \Rightarrow 5x^2+5y^2={\left(y-2x+10\right)}^2 \\ \Rightarrow x^2+4y^2+4xy+40x-20y-100=0 \\ \Rightarrow {\left(x+2y\right)}^2+40x-20y-100=0 \end{gathered}$$

Answer:

(i) Given:
  y² = 8x

On comparing the given equation with $y^2=4ax$:
$4a=8\Rightarrow a=2$

∴ Vertex = (0, 0)

Focus = (a, 0) = (2, 0)

Equation of the directrix:
x = −a
i.e. x = −2

Axis = y = 0

Length of the latus rectum = 4a = 8 units
 

(ii) Given:
  4x² + y = 0

$\Rightarrow \frac{-y}{4}=x^2$

On comparing the given equation with $x^2=-4ay$:

$4a=\frac{1}{4}\Rightarrow a=\frac{1}{16}$

∴ Vertex = (0, 0)

Focus = (0, −a) = $\left(0,\frac{-1}{16}\right)$

Equation of the directrix:
y = a
i.e. $y=\frac{1}{16}$

Axis = x = 0

Length of the latus rectum = 4a = $\frac{1}{4}$ units

(iii) Given: 
y² − 4y − 3x + 1 = 0

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2-4-3x+1=0 \\ \Rightarrow {\left(y-2\right)}^2=3\left(x+1\right) \\ \Rightarrow {\left(y-2\right)}^2=3\left(x-\left(-1\right)\right) \end{gathered}$$

Let $Y=y-2$, $X=x+1$
Then, we have:
$Y^2=3X$

Comparing the given equation with $Y^2=4aX$:

$4a=3\Rightarrow a=\frac{3}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=-1, y=2\right)$

Focus = (X = a, Y = 0) = $\left(x+1=\frac{3}{4}, y-2=0\right)=\left(x=\frac{-1}{4}, y=2\right)$

Equation of the directrix:
X = −a
i.e.  $x+1=\frac{-3}{4}\Rightarrow x=\frac{-7}{4}$

Axis = Y = 0
i.e. $y-2=0\Rightarrow y=2$

Length of the latus rectum = 4a = 3 units

(iv) Given:
y² − 4y + 4x = 0

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2-4+4x=0 \\ \Rightarrow {\left(y-2\right)}^2=-4\left(x-1\right) \end{gathered}$$

Let $Y=y-2$, $X=x-1$
Then, we have:
$Y^2=-4X$

Comparing the given equation with $Y^2=-4aX$:

$4a=4\Rightarrow a=1$

∴ Vertex = (X = 0, Y = 0) = $\left(x=1, y=2\right)$

Focus = (X = −a, Y = 0) = $\left(x-1=-1, y-2=0\right)=\left(x=0, y=2\right)$

Equation of the directrix:
X = a
i.e. $x-1=1\Rightarrow x=2$

Axis = Y = 0
i.e. $y-2=0\Rightarrow y=2$

Length of the latus rectum = 4a = 4 units

(v) Given:
y² + 4y + 4x −3 = 0

$$\begin{gathered} \Rightarrow {\left(y+2\right)}^2-4+4x-3=0 \\ \Rightarrow {\left(y+2\right)}^2=-4\left(x-\frac{7}{4}\right) \end{gathered}$$

Let $Y=y+2$, $X=x-\frac{7}{4}$
Then, we have:
$Y^2=-4X$

Comparing the given equation with $Y^2=-4aX$:
$4a=4\Rightarrow a=1$

∴ Vertex = (X = 0, Y = 0) = $\left(x=\frac{7}{4}, y=-2\right)$

Focus = (X = −a, Y = 0) = $\left(x-\frac{7}{4}=-1, y+2=0\right)=\left(x=\frac{3}{4}, y=-2\right)$

Equation of the directrix:
X = a
i.e. $x-\frac{7}{4}=1\Rightarrow x=\frac{11}{4}$

Axis = Y = 0
i.e. $y+2=0\Rightarrow y=-2$

Length of the latus rectum = 4a = 4 units

(vi) Given:
y² = 8x + 8y
$\Rightarrow {\left(y-4\right)}^2=8\left(x+2\right)$

Putting $Y=y-4$, $X=x+2$:
$Y^2=8X$


On comparing the given equation with $Y^2=4aX$:
$4a=8\Rightarrow a=2$

∴ Vertex = (X = 0, Y = 0) = $\left(x=-2, y=4\right)$

Focus = (X = a, Y = 0) = $\left(x+2=2, y-4=0\right)=\left(x=0, y=4\right)$
Equation of the directrix:
X = −a
i.e. $x+2=-2\Rightarrow x+4=0$
Axis = Y = 0
i.e. $y-4=0\Rightarrow y=4$
Length of the latus rectum = 4a = 8

(vii) Given:
4(y − 1)² = − 7 (x − 3)
$\Rightarrow {\left(y-1\right)}^2=\frac{-7}{4}\left(x-3\right)$

Let $Y=y-1$, $X=x-3$
Then, we have:
$Y^2=\frac{-7}{4}X$

Comparing the given equation with $Y^2=-4aX$:
$4a=\frac{7}{4}\Rightarrow a=\frac{7}{16}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=3, y=1\right)$

Focus = (X = −a, Y = 0) = $\left(x-3=\frac{-7}{16}, y-1=0\right)=\left(x=\frac{41}{16}, y=1\right)$

Equation of the directrix:
X = a
i.e. $x-3=\frac{7}{16}\Rightarrow x=\frac{55}{16}$

Axis = Y = 0
i.e. $y-1=0 \Rightarrow y = 1$

Length of the latus rectum = 4a = $\frac{7}{4}$ units

(viii) Given:
  y ² = 5x − 4y − 9

$$\begin{gathered} \Rightarrow y^2+4y=5x-9 \\ \Rightarrow {\left(y+2\right)}^2=5x-5=5\left(x-1\right) \end{gathered}$$

Putting $Y=y+2$, $X=x-1$:
$Y^2=5X$

Comparing the given equation with $Y^2=4aX$:
$4a=5\Rightarrow a=\frac{5}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=1, y=-2\right)$

Focus = (X = a, Y = 0) = $\left(x-1=\frac{5}{4}, y+2=0\right)=\left(x=\frac{9}{4}, y=-2\right)$

Equation of the directrix:
X = −a
i.e. $x-1=\frac{-5}{4}\Rightarrow x=\frac{-1}{4}$

Axis = Y = 0
i.e. $y+2=0\Rightarrow y=-2$

Length of the latus rectum = 4a = 5 units

(ix) Given:
x² = 6x−y−14
$$\begin{gathered} \Rightarrow {\left(x-3\right)}^2=-y-14+9 \\ \Rightarrow {\left(x-3\right)}^2=-y-5=-\left(y+5\right) \end{gathered}$$

Let $Y=y+5$, $X=x-3$
Then, we have:
$X^2=-Y$

Comparing the given equation with $X^2=-4aY$:
$4a=1\Rightarrow a=\frac{1}{4}$

∴ Vertex = (X = 0, Y = 0) = $\left(x=3, y=-5\right)$

Focus = (X = 0, Y = −a) = $\left(x-3=0, y+5=\frac{-1}{4}\right)=\left(x=3, y=\frac{-21}{4}\right)$

Equation of the directrix:
Y = a

i.e. $y+5=\frac{1}{4}\Rightarrow y=\frac{-19}{4}$

Axis = X = 0
i.e. $x-3=0\Rightarrow x=3$

Length of the latus rectum = 4a = 1 units

Answer:

The given equation of the parabola is y² = 4px.

Let PQ be the double ordinate of length 8p of the parabola $y^2=4px$.

Then, we have:
PR = RQ = 4p

Let AR = x₁
Then, the coordinates of P and Q are $\left(x_1,4p\right)$ and $\left(x_1,-4p\right)$, respectively.

Now, P lies on $y^2=4px$.

∴ ${\left(4p\right)}^2=4p{x}_1$
$\Rightarrow x_1=4p$

So, the coordinates of P and Q are $\left(4p, 4p\right)$ and $\left(4p,-4p\right)$, respectively.

The coordinates of A are (0, 0).

$$\begin{gathered} \therefore m_1=\mathrm{slope} \mathrm{of}\mathit{ }AP=\frac{4p-0}{4p-0}=1 \\ \mathrm{And}, m_2=\mathrm{slope} \mathrm{of}\mathit{ }AQ=\frac{\left(-4p-0\right)}{4p-0}=-1 \end{gathered}$$

Now, $m_1{m}_2=-1$

Thus, AP is perpendicular to AQ.

Hence, the lines from the vertex to its extremities are at right angles.

Answer:

The given equation of the parabola is x² = 12y.

On comparing the given equation with $x^2=4ay$:
a = 3



Required area = $\frac{1}{2}\left(LL\text{'}\times OS\right)=\frac{1}{2}\times 12\times 3=18 \mathrm{square} \mathrm{units}$
                                       

Answer:

The given equation of the directrix is 3x − 4y = 2.                        (1)

∴ Slope of the directrix = $\frac{-3}{-4}=\frac{3}{4}$

Also, the axis is perpendicular to the directrix.
∴ Slope of the axis = $\frac{-4}{3}$

The focus lies on the axis of the parabola.
∴ Equation of the axis:
$\left(y-3\right)=\frac{-4}{3}\left(x-3\right)$
⇒ $\left(3y-9\right)=-4x+12$

⇒ $3y+4x-21=0$                         (2)


Solving equations (1) and (2):

$x=\frac{18}{5}, y=\frac{11}{5}$

Therefore, the intersection point of the axis and directrix is $\left(\frac{18}{5},\frac{11}{5}\right)$.
Also, length of the latus rectum = 2 (Length of the perpendicular from the focus on the directrix)
                                             =$2\left|\frac{3\left(3\right)+\left(-4\right)3-2}{\sqrt{16+9}}\right|=2\left|\frac{-5}{\sqrt{16+9}}\right|=2$ square units

Answer:

Putting x = 3y in the given equation of the parabola:
$$\begin{gathered} 9y^2=9y \\ \Rightarrow 9y\left(y-1\right)=0 \\ \Rightarrow y=0,1 \end{gathered}$$
 
At y = 0, x = 0
At y = 1, x = 3

Therefore, at (1, 3), the abscissa is three times that of the ordinate.

Answer:

CASE I:
Let the equation of the required parabola be $y^2=4ax$                         (1)

Since (1) passes through (2, 3), we have:

 $9=4a\left(2\right)\Rightarrow a=\frac{9}{8}$

Thus, the required equation is $y^2=\frac{4\left(9\right)x}{8}$, i.e. $2y^2=9x$.

CASE II:
Let the equation of the required parabola be $y^2=-4ax$                   (2)

Since (2) passes through (2, 3), we have:

 $9=-4a\left(2\right)\Rightarrow a=\frac{-9}{8}$

Thus, the required equation is $y^2=\frac{-4\left(-9\right)x}{8}$, i.e. $2y^2=9x$.

Hence, in either case, the required equation of the parabola is $2y^2=9x$.

Answer:

Let the equation of the directrix be y = a.      

Comparing with y = 2:
a = 2

Equation of the parabola with directrix y =a is $x^2=-4ay$.

Hence, the required equation of the parabola is $x^2=-8y$.

Answer:

Given:
The vertex and the focus of the parabola are (3, 2) and (5, 2), respectively.

∴ Slope of the axis of the parabola = 0

 Slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+5}{2}=3,\frac{s+2}{2}=2 \\ \Rightarrow r=1, s=2 \end{gathered}$$

Required equation of the directrix is $x-1=0$, which can be rewritten as x = 1.

Let P (x, y) be any point on the parabola whose focus is S (5, 2) and the directrix is x =1.

Draw PM perpendicular to x  = 1.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-5\right)}^2+{\left(y-2\right)}^2={\left(\frac{x-1}{1}\right)}^2 \\ \Rightarrow x^2+25-10x+y^2+4-4y=x^2+1-2x \\ \Rightarrow 25-10x+y^2+4-4y-1+2x=0 \\ \Rightarrow y^2-4y-8x+28=0 \end{gathered}$$

Answer:

Let X'OX be the bridge and PAQ be the suspension cable.
The suspension cable forms a parabola with the vertex at (0, 6).

Let the equation of the parabola formed by the suspension cable be ${\left(x-0\right)}^2=4a\left(y-6\right)$.                                 (1)

It passes through P (−50, 30) and Q (50, 30).

∴ $2500=4a\left(30-6\right)$
⇒ $4a=\frac{2500}{24}$

Putting the value of 4a in equation (1):

$x^2=\frac{2500}{24}\left(y-6\right)$                                     (2)

Let LM be the supporting wire attached at M, which is 18 m from the mid-point (O) of the bridge.

Let the coordinates of L be (18, l).
It lies on the parabola (2).

∴ ${18}^2=\frac{2500}{24}\left(l-6\right)$
$\Rightarrow l=9.11 \mathrm{m}$

Hence, the length of the supporting wire attached to the roadway 18 m from the middle is 9.11 m.

Answer:

Let A and B be points on the parabola y² = 6x and OA, OB be the lines joining the vertex O to the points A and B whose abscissa are 24.



Now,
y² = 6 × 24 = 144
⇒ y = ± 12
Therefore the coordinates of the points A and B are (24, 12) and (24, –12) respectively.
Hence the lines are given by
  $$\begin{gathered} y-0=\pm \frac{12-0}{24-0}(x-0) \\ \Rightarrow \pm 2y=x \end{gathered}$$

Answer:

We have y² = 8x
⇒ y² = 4(2)x
Comparing it with the general equation of parabola y² = 4ax, we will get a = 2
Let the required point be (x₁, y₁)
Now, Focal distance = 4
⇒ x₁ + a = 4
⇒ x₁ + 2 = 4
⇒ x₁ = 2
Now, the point will satisfy the equation of parabola
∴ (y₁)² = 8(2) = 16
⇒ y₁ = ± 4
Hence, the coordiantes of the points are (2, 4) and (2, −4).

Answer:

Let the coordinates of the point on the parabola be B (x₁, y₁).
Let BO be the line segment
In right triangle AOB
$$\begin{gathered} \mathrm{cos\theta }=\frac{\mathrm{AO}}{\mathrm{OB}} \mathrm{and} \mathrm{sin\theta }=\frac{\mathrm{AB}}{\mathrm{OB}} \\ \Rightarrow \mathrm{cos\theta }=\frac{x_1}{\mathrm{OB}} \mathrm{and} \mathrm{sin\theta }=\frac{y_1}{\mathrm{OB}} \end{gathered}$$
∴ x₁ = OBcosθ and y₁ = OBsinθ
Now, the curve is passing through (x₁, y₁)
∴ (y₁)² = 4a(x₁)
⇒( OBsinθ)² = 4a(OBcosθ)
$$\begin{gathered} \Rightarrow {\mathrm{OB}}^2{\sin }^2\mathrm{\theta }=4a\mathrm{OBcos\theta } \\ \Rightarrow \mathrm{OB}=\frac{4a\mathrm{cos\theta }}{{\sin }^2\mathrm{\theta }}=4a\mathrm{cosec}\theta .\cot \theta \end{gathered}$$

Answer:

As the vertex and focus lie on y-axis, so y-axis is the axis of the parabola.
If the directrix meets the axis of the parabola at point Z, the AZ = AF = 2
OZ = OF + AZ + FA = 2 + 2 + 2 = 6
So, the equation of the directrix is y = 6
i.e., y − 6 = 0
Let P(x, y) be any point in the plane of the focus and directrix and MP be the perpendicular
distance from P to the directrix, then P lies on parabola iff FP = MP

Answer:

We have y² = 4x
Substituting the value of  y = mx + 1 in y² = 4x, we get
(mx + 1)² = 4x
⇒ m²x² + 2mx + 1 = 4x
⇒ m²x² + (2m − 4)x + 1 = 0                       .....(1)
Since, a tangent touches the curve at a point, the roots of (1) must be equal.
∴ D = 0
⇒ (2m − 4)² − 4m² = 0
⇒ 4m² −16m + 16 − 4m² = 0
⇒ m = 1
 

Answer:

(a)  (5/4, 1)

Given:
The equation of the parabola is y² − x − 2y + 2 = 0.
  $$\begin{gathered} \Rightarrow {\left(y-1\right)}^2-1=\left(x-2\right) \\ \Rightarrow {\left(y-1\right)}^2=x-1 \end{gathered}$$

Let $X = x-1, Y = y-1$

∴ $Y^2=X$

Comparing with $Y^2=4aX$:
$a=\frac{1}{4}$

Focus = $\left(X=a, Y=0\right)=\left(X=\frac{1}{4}, Y=0\right)=\left(x=\frac{1}{4}+1, y=1\right)=\left(x=\frac{5}{4}, y=1\right)$

Hence, the focus is at (5/4, 1).

Answer:

(b) (a, −a)

Given:
The equation of the parabola is (y + a)² = 8a (x − a).
 
Putting $X= x-a, Y=y+a$:

$Y^2=8aX$

Vertex = $\left(X=0, Y=0\right)=\left(x-a=0, y+a=0\right)=\left(x=a, y=-a\right)$

Hence, the vertex is at (a, −a).

Answer:

(c)  (−1, 2)

Given:
The focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0.

The slope of the line perpendicular to x + y − 3 = 0 is 1.

The axis of the parabola is perpendicular to the directrix and passes through the focus.

∴ Equation of the axis of the parabola = $y-1=1\left(x+2\right)$             (1)

Intersection point of the directrix and the axis is the intersection point of (1) and x + y − 3 = 0.

Let the intersection point be K.

Therefore, the coordinates of K will be (0, 3).

Let (h, k) be the coordinates of the vertex, which is the mid-point of the segment joining K and the focus.

$$\begin{gathered} \therefore h=\frac{0-2}{2}, k=\frac{3+1}{2} \\ h=-1, k=2 \end{gathered}$$

Hence, the coordinates of the vertex are (−1, 2).

Answer:

(b) x² − 2xy + y² + 6ax + 10ay − 7a² = 0

Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.

The slope of the line perpendicular to x + y = 3a is 1.

The axis of the parabola is perpendicular to the directrix and passes through the vertex.

∴ Equation of the axis of the parabola = $y-0=1\left(x-a\right)$                           (1)

Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.

Let the intersection point be K.

Therefore, the coordinates of K are $\left(2a, a\right)$.

The vertex is the mid-point of the segment joining K and the focus (h, k).
∴ $$\begin{gathered} a=\frac{2a+h}{2}, 0=\frac{a+k}{2} \\ h=0, k=-a \end{gathered}$$

Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y = 3a.

Draw PM perpendicular to x + y = 3a.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y+a\right)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2 \\ \Rightarrow x^2+{\left(y+a\right)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2 \\ \Rightarrow 2x^2+2y^2+2a^2+4ay=x^2+y^2+9a^2+2xy-6ax-6ay \\ \Rightarrow x^2+y^2-7a^2+10ay+6ax-2xy=0 \end{gathered}$$

Answer:

(a) x = 0

Given:
x = t² + 1         (1)
y = 2t + 1         (2)

From (1) and (2):
$x={\left(\frac{y-1}{2}\right)}^2+1$

On simplifying:
${\left(y-1\right)}^2=4\left(x-1\right)$

Let $Y=y-1 \mathrm{and} X=x-1$

∴ $Y^2=4X$

Comparing it with y² = 4ax:
a = 1

Therefore, the equation of the directrix is X = −a , i.e. $x-1=-1\Rightarrow x=0$.

Answer:

(a) 3x + 2y + 14 = 0

Given:
The vertex and the focus of a parabola are (−1, 1) and (2, 3), respectively.

∴ Slope of the axis of the parabola = $\frac{3-1}{2+1}=\frac{2}{3}$

 Slope of the directrix = $\frac{-3}{2}$

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+2}{2}=-1,\frac{s+3}{2}=1 \\ \Rightarrow r=-4, s=-1 \end{gathered}$$

Equation of the directrix:
    $\left(y+1\right)=\frac{-3}{2}\left(x+4\right)$
$\Rightarrow 3x+2y+14=0$

Answer:

(c) parabola



Suppose PQ is a double ordinate of the parabola $y^2=4ax$.

Let R and S be the points of trisection of the double ordinates.

Let $\left(h,k\right)$ be the coordinates of R.

Then, we have:
OL = h  and RL = k

$$\begin{gathered} \therefore RS=RL+LS=k+k=2k \\ \Rightarrow PR=RS=SQ=2k \\ \Rightarrow LP=LR+RP=k+2k=3k \end{gathered}$$

Thus, the coordinates of P are $\left(h, 3k\right)$, which lie on $y^2=4ax$.

∴ $9k^2=4ah$

Hence, the locus of the point (h, k) is $9y^2=4ax$,  i.e.  $y^2=\left(\frac{4a}{9}\right)x$, which represents a parabola.

Answer:

(a) x + 2y = 4

Given:
The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively.

∴ Slope of the axis of the parabola = $\frac{6-4}{2-1}=2$

Slope of the directrix = $\frac{-1}{2}$

Let the directrix intersect the axis at K (r, s).

∴ $$\begin{gathered} \frac{r+2}{2}=1,\frac{s+6}{2}=4 \\ \Rightarrow r=0, s=2 \end{gathered}$$

Equation of the directrix:

    $\left(y-2\right)=\frac{-1}{2}\left(x-0\right)$
$\Rightarrow$ x + 2y = 4

Answer:

 (b) 1/2

Given:
The vertex and the focus of a parabola are V and S, respectively.

The given equation of parabola can be rewritten as follows:
   ${\left(y+3\right)}^2-9+5+2x=0$
$$\begin{gathered} \Rightarrow {\left(y+3\right)}^2+2x=4 \\ \Rightarrow {\left(y+3\right)}^2=4-2x \\ \Rightarrow {\left(y+3\right)}^2=-2\left(x-2\right) \end{gathered}$$

Let $Y=y+3, X=x-2$

Then, the equation of parabola becomes $Y^2=-2X$.

Vertex = $\left(X=0, Y=0\right)=\left(x-2=0, y+3=0\right)=\left(x=2, y=-3\right)$

Comparing with y² = 4ax:

$4a=2\Rightarrow a=\frac{1}{2}$

Focus = $\left(X=\frac{-1}{2}, Y=0\right)=\left(x-2=\frac{-1}{2}, y+3=0\right)=\left(x=\frac{3}{2}, y=-3\right)$

⇒ SV = $\sqrt{{\left(2-\frac{3}{2}\right)}^2+{\left(-3+3\right)}^2}=\frac{1}{2}$

Answer:

(c)  y = −1

Given:
x² − 4x − 8y + 12 = 0
$$\begin{gathered} \Rightarrow {\left(x-2\right)}^2-4-8y+12=0 \\ \Rightarrow {\left(x-2\right)}^2=8y-8 \\ \Rightarrow {\left(x-2\right)}^2=8\left(y-1\right) \end{gathered}$$

Putting X = x − 2, Y = y − 1:

$X^2=8Y$

Comparing with $X^2=4aY$:
a = 2

Equation of the directrix:
     $Y=-a$
⇒ $Y=-2$
$$\begin{gathered} \Rightarrow y-1=-2 \\ \Rightarrow y=-2+1 \\ \Rightarrow y=-1 \end{gathered}$$

Answer:

(a) x² + y² − 2xy + 8x + 8y − 16 = 0

Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is x + y = 4.

Draw PM perpendicular to x + y = 4.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(\frac{x+y-4}{\sqrt{2}}\right)}^2 \\ \Rightarrow x^2+y^2={\left(\frac{x+y-4}{\sqrt{2}}\right)}^2 \\ \Rightarrow 2x^2+2y^2=x^2+y^2+16+2xy-8x-8y \\ \Rightarrow x^2+y^2-2xy+8x+8y-16=0 \end{gathered}$$

Answer:

(c) (−1, 2)

Let the coordinates of P and Q be $\left(a{t_1}^2, 2a{t}_1\right)$ and $\left(a{t_2}^2, 2a{t}_2\right)$, respectively.

Slope of PQ = $\frac{2a{t}_2-2a{t}_1}{a{t_2}^2-a{t_1}^2}$               ......(1)

But, the slope of PQ is equal to the slope of 2x − y + 4 = 0.

∴ Slope of PQ = $\frac{-2}{-1}=2$

From (1),

$\frac{2a{t}_2-2a{t}_1}{a{t_2}^2-a{t_1}^2}=2$                                .....(2)

Putting 4a = 8,
a = 2

∴ Focus of the given parabola = (a, 0) = $\left(2, 0\right)$.

Using equation (2):

$\frac{4\left(t_2-t_1\right)}{2\left({t_2}^2-{t_1}^2\right)}=2$
$\frac{\left(t_2-t_1\right)}{\left({t_2}^2-{t_1}^2\right)}=1$
$\Rightarrow t_1+t_2=1$
As, points P and Q lie on 2x-y+4=0

$$\begin{gathered} \Rightarrow P(a{t_1}^2, 2a{t}_1) \mathrm{or} P(2{t_1}^2, 4t_1) \mathrm{lie} \mathrm{on} \mathrm{line} 2x-y+4=0 \\ \Rightarrow 2\left(2{t_1}^2\right)-\left(4t_1\right)+4=0 \\ \Rightarrow {t_1}^2-t_1+1=0 ...\left(3\right) \\ Also, Q(a{t_2}^2, 2a{t}_2) \mathrm{or} P(2{t_2}^2, 4t_2) \mathrm{lie} \mathrm{on} \mathrm{line} 2x-y+4=0 \\ \Rightarrow 2\left(2{t_2}^2\right)-\left(4t_2\right)+4=0 \\ \Rightarrow {t_2}^2-t_2+1=0 ...\left(4\right) \\ \mathrm{Adding} \left(3\right) \mathrm{and} \left(4\right), \mathrm{we} \mathrm{get}, \\ \Rightarrow {t_1}^2-t_1+1+{t_2}^2-t_2+1=0 \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)-\left(t_1+t_2\right)+2=0 \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)-1+2=0 \left[t_1+t_2=1, \mathrm{proved} \mathrm{above}\right] \\ \Rightarrow \left({t_1}^2+{t_2}^2\right)=-1 \end{gathered}$$

Let $\left(x_1, y_1\right)$ be the mid-point of PQ.

Then, we have:
$y_1=\frac{2a{t}_2+2a{t}_1}{2}=2\left(t_1+t_2\right)=2$
And, $x_1=\frac{a{t_1}^2+a{t_2}^2}{2}={t_1}^2+{t_2}^2=-1$

⇒ $\left(x_1, y_1\right)=\left(-1, 2\right)$

Answer:

(a) $4\sqrt{2}a$



Let OP be the chord.

Let the coordinates of P be $\left(x_1, y_1\right)$.

From the figure, we have:

$O{P}^2={x_1}^2+{y_1}^2$                   (1)
And, $\tan \frac{\mathrm{\pi }}{4}=\frac{y_1}{x_1}$
$\Rightarrow x_1=y_1$                           (2)

Also, $\left(x_1, y_1\right)$ lies on the parabola.

∴ ${y_1}^2=4a{x}_1$                    (3)

Using (2) and (3):
${x_1}^2=4a{x}_1\Rightarrow x_1=4a$       (4)

∴ From (4), (1) and (2), we have:
$$\begin{gathered} O{P}^2={\left(4a\right)}^2+{\left(4a\right)}^2=32a^2 \\ \Rightarrow OP=4\sqrt{2}a \end{gathered}$$

Therefore, the length of the chord is $4\sqrt{2}a \mathrm{units}$.

Answer:

(b) a parabola
Comparing the given equation with ax² + by² + 2hxy + 2gx+2fy + c =0, we get:
$a=16, b=1, h=4$

We have: $h^2=16=ab$

Thus, the given equation represents a parabola.

Answer:

(c) 8

Given:
y² + 8x − 2y + 17 = 0
$$\begin{gathered} \Rightarrow {\left(y-1\right)}^2-1+8x+17=0 \\ \Rightarrow {\left(y-1\right)}^2+8x+16=0 \\ \Rightarrow {\left(y-1\right)}^2=-8\left(x+2\right) \end{gathered}$$

Let $X=x+2, Y=y-1$

∴ $Y^2=-8X$

Comparing with y²=4ax:
a = 2

Length of the latus rectum = 4a = 8 units

Answer:

(a) (−4, 1)

Given:
x² + 8x + 12y + 4 = 0
$$\begin{gathered} \Rightarrow {\left(x+4\right)}^2-16+12y+4=0 \\ \Rightarrow {\left(x+4\right)}^2+12y-12=0 \\ \Rightarrow {\left(x+4\right)}^2=-12\left(y-1\right) \end{gathered}$$

Let $X=x+4, Y=y-1$

∴ $X^2=-12Y$

Vertex = $\left(X=0,Y=0\right)=\left(x+4=0,y-1=0\right)=\left(x=-4,y=1\right)$

Hence, the vertex is at (−4, 1).

Answer:

(a) (1, 2)

Given:
(y − 2)² = 16 (x − 1)

Let $X=x-1, Y=y-2$

∴ $Y^2=16X$

Vertex = $\left(X=0, Y=0\right)=\left(x-1=0, y-2=0\right)=\left(x=1, y=2\right)$

Hence, the vertex is at (1, 2).

Answer:

(c) 1/2

Given:
4y² + 2x − 20y + 17 = 0

$$\begin{gathered} \Rightarrow y^2+\frac{x}{2}-5y+\frac{17}{4}=0 \\ \Rightarrow {\left(y-\frac{5}{2}\right)}^2+\frac{x}{2}-2=0 \\ \Rightarrow {\left(y-\frac{5}{2}\right)}^2=-1\left(\frac{x}{2}-2\right) \\ \Rightarrow {\left(y-\frac{5}{2}\right)}^2=\frac{-1}{2}\left(x-4\right) \end{gathered}$$

Let $X=x-4, Y=y-\frac{5}{2}$

$\therefore Y^2=\frac{-X}{2}$

∴ Length of the latus rectum = 4a = $\frac{1}{2}$ units