1 Marks Question

Question 11 Mark

Write the axis of symmetry of the parabola y² = x.

Answer

The axis of symmetry of the parabola y² = x is x-axis.

Question 21 Mark

Write the length of the chord of the parabola y² = 4ax which passes through the vertex and is inclined to the axis at $\frac{\pi}{4}.$

Answer

Let A be the vertex of the parabola. Then cooridnates of A are (0, 0).
Suppose AP is a chord that is inclined to an angle of $\frac{\pi}{4}$ radians to the X axis. Let M be the point where the perpendicular from P intersects the X axis.
Let AP = l
Then,
$\frac{\text{AM}}{\text{l}}=\cos\frac{\pi}{4}$
$\Rightarrow\ \text{AM}=\text{l}\times\frac{1}{\sqrt2}$
$\Rightarrow\ \text{AM}=\frac{\text{l}}{\sqrt2}$
$\text{and }\frac{\text{PM}}{\text{l}}=\sin\frac{\pi}{4}$
$\Rightarrow\ \text{PM}=\text{l}\times\frac{1}{\sqrt2}=\frac{\text{l}}{\sqrt2}$
So, that coordinates of P are $\Big(\frac{\text{l}}{\sqrt2}, \frac{\text{l}}{\sqrt2}\Big).$
Since, P lies on y² = 4ax
$\therefore\ \Big(\frac{\text{l}}{\sqrt2}\Big)^2=4\text{a}\times\frac{\text{l}}{\sqrt2}$
$\Rightarrow\ \frac{\text{l}^2}{2}=4\text{a}\frac{\text{l}}{\sqrt2}$
$\Rightarrow\ \text{l}=\frac{8\text{a}}{\sqrt2}$
$\Rightarrow\ \text{l}=\frac{4\times\sqrt2\times\sqrt2\text{a}}{\sqrt2}$
$=\ 4\sqrt2\text{a}$
$\Rightarrow\ \text{l}=4\sqrt2\text{a}$

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Question 31 Mark

Write the equation of the parabola whose vertex is at (-3, 0) and the directrix is x + 5 = 0.

Answer

The general equation of the parabola is (y - k)² = 4a(x - h)
Here, the (h, k) = (-3, 0)
Now, the directrix is given by
x = h - a
⇒ -5 = -3 - a [$\because$ x + 5 = 0 ⇒ x = -5]
⇒ a = 2
Hence, the equation is given by
(y - 0)² = 4(2)(x + 3)
⇒ y² = 8 (x + 3)

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Question 41 Mark

Write the equation of the parabola with focus (0, 0) and directrix x + y - 4 = 0.

Answer

Let P(x, y) be any point on the parabola whose focus is S(0, 0) and the directrix
x + y - 4 = 0
Draw PM perpendicular from P(x, y) on the directrix
x + y - 4 = 0
Then by definition,
SP = PM
⇒ SP² = PM²
$\Rightarrow\ (\text{x} - \text{0})^2 + (\text{y} - \text{0})^2 =\Big[\frac{\text{x+y}-4}{\sqrt{1^2+1^2}}\Big]^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=\frac{(\text{x+y}-4)^2}{(\sqrt2)^2}$
⇒ 2x² + 2y² = x² + y² + (-4)² + 2xy + 2 × y × (-4) + 2 × (-4) × x
⇒ 2x² + 2y² = x² + y² + 16 + 2xy - 8y - 8x
⇒ 2x² - x² + 2y² - y² - 2xy + 8x + 8y -16 = 0
⇒ x² + y² - 2xy + 8x + 8y - 16 = 0.

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Question 51 Mark

If the coordinates of the vertex and focus of a parabola are (-1, 1) and (2, 3) respectively, then write the equation of its directrix.

Answer

The equation of line posses through vertex and focus of a parabola is

$\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{x}-\text{x}_1}$

$\Rightarrow\ \frac{3-1}{2-(-1)}=\frac{\text{y}-1}{\text{x}-(-1)}$ [$\because$ Focus: (2, 3) and vertex: (-1, 1)]

$\Rightarrow\ \frac{2}{3}=\frac{\text{y}-1}{\text{x}+1}$

⇒ 2x + 2 = 3y - 3

⇒ 3y - 2x - 3 - 2 = 0

⇒ 3y - 2x - 5 = 0 ...(i)

The equation of $\bot$ line to

3y - 2x - 5 = 0 is

$2\text{y} + 3\text{x} + \lambda = 0 ...(\text{ii})$

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

Then(-1, 1) is the mid-point of the line segment joining (2, 3) and (x_1,y₁).

$\therefore\ \frac{\text{x}_1+2}{2}=-1\text{ and }\frac{\text{y}_1+3}{2}=1$

⇒ x₁ + 2 = -2 and y₁ + 3 = 2

⇒ x₁ = -4 and y₁ = -1

Thus, the directrix meets the axis at (-4, -1).

$\therefore$ The prependicular line $2\text{y} + 3\text{x} +\lambda = 0$ posses through (-4, -1).

$\therefore\ 2(-1) + 3(-4) + \lambda = 0$

$\Rightarrow\ - 2 - 12 + \lambda = 0$

$\Rightarrow\ \lambda=14$

Putting $\lambda=14$ in equation (ii), we get

2y + 3x + 14 = 0

Hence, the required equation of directrix is 2y + 3x + 14 = 0.

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Question 61 Mark

If the parabola y² = 4ax passes through the point (3, 2), then find the length of its latusrectum.

Answer

We have y² = 4ax
Since, the parabola is passing through the point (3, 2)
Hence, it will satisfy the equation of the parabola.
$\therefore$ 2² = 4(a)(3)
$\Rightarrow\ \text{a}=\frac{1}{3}$
Lenth of the latus ractum is given by,
4a
$=\ 4\times\frac{1}{3}$
$=\ \frac{4}{3}$

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Question 71 Mark

Ifb and care lengths of the segments of any focal chord of the parabola y² = 4ax, then write the length of its latus-rectum.

Answer

Let S(a, 0) be the focus of the given parabola.
Let the end points of the focal chord be $\text{P}(\text{at}^2, 2\text{at})$ and $\text{Q}\Big(\frac{\text{a}}{\text{t}^2},\ \frac{-2\text{a}}{\text{t}}\Big).$
SP and SQ are segments of the focal chord with lengths b and c, respectively.
$\therefore$ SP = b, SQ = c
Also, $\text{SP}=\sqrt{(\text{a}-\text{at}^2)+4\text{a}^2\text{t}^2}=\text{a}(1+\text{t}^2)$
And, $\text{SQ}=\sqrt{\Big(\text{a}-\frac{\text{a}}{\text{t}^2}\Big)+\frac{4\text{a}^2}{\text{t}^2}}=\text{a}\Big(1+\frac{1}{\text{t}^2}\Big)$
Now, we have:
$\frac{1}{\text{SP}}+\frac{1}{\text{SQ}}=\frac{1}{\text{a}(1+\text{t}^2)}+\frac{\text{t}^2}{\text{a}(1+\text{t}^2)}=\frac{1}{\text{a}}$
$\Rightarrow\ \frac{1}{\text{b}}+\frac{1}{\text{c}}=\frac{1}{\text{a}}$
$\Rightarrow\ \frac{\text{b+c}}{\text{bc}}=\frac{1}{\text{a}}$
$\Rightarrow\ \text{a}=\frac{\text{bc}}{\text{b+c}}$
$\therefore$ Length of the latus rectum $=4\text{a}=\frac{4\text{bc}}{\text{b+c}}$

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Question 81 Mark

Write the distance between the vertex and focus of the parabola y² + 6y + 2x + 5 = 0.

Answer

We have,
y²+ 6y = -2x - 5
⇒ y² + 2 × y × 3 + 9 = -2x - 5 + 9
⇒ (y + 3)² = -2x + 4
⇒ (y + 3)² = -2(x - 2) ...(i)
Shifting the origin to point (2, - 3) without rotating the axes and denoting the new coordinates w.r.t these axes by x and y.
we have
x = x + 2, y = y - 3 ...(ii)
Using these relation, equation (i) reduces to
y² = -2x ...(iii)
This is of the form y² = -4ax. on comparing, we get
4a = 2
$\Rightarrow\ \text{a}=\frac{2}{4}=\frac{1}{2}$
$\therefore$ The distance between the vertex and focus of the parabola is $\frac{1}{2}.$

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Question 91 Mark

Write the equation of the directrix of the parabola x² - 4x - 8y + 12 = 0.

Answer

The given system of equation is
x² - 4x - 8y +12 =0
⇒ x² - 4x = 8y - 12
⇒ x² - 2 × x × 2 + 4 = 8y - 12 + 4
⇒ (x - 2)² = 8y - 8
⇒ (x - 2)² = 8(y - 1) ...(i)
Shifting the origin to point (2, 1) without rotating the axes and denoting the new coordinates w.r.t these axes by x and y, we have
x = x + 2, y = y - 3 ...(ii)
Using these relation, equation (i) reduces to
y² = 8y ...(iii)
This is of the form x² = 4ay. on comparing, we get
4a = 8
⇒ a = 2
$\therefore$ equation of the directrix of the parabola w.r.t new axes is
y = -2
$\therefore$ y = -2 + 1 [Using equation (ii)]
⇒ y = -1
⇒ y + 1 = 0
$\therefore$ equation of the directrix of the parabola w.r.t old axes is y + 1 = 0.

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Question 101 Mark

Write the coordinates of the vertex of the parabola whose focus is at(-2, 1) and directrix is the line x + y - 3 = 0.

Answer

The equation of directrix is
x + y - 3 = 0 ...(i)
⇒ y = -x + 3
$\therefore$ slope of line = m₁ = -1
$\therefore$ slope of line $=\ \text{m}_2=\frac{-1}{\text{m}_1}=\frac{-1}{-1}=1$
The equation of line posess through (-2, 1) with slope 1 is
y - 1 = 1 [x - (-2)] [$\because$ y - y₀ = m(x - x₀)]
⇒ y - 1 = x + 2
⇒ y - x = 3
⇒ y - x - 3 = 0 ...(ii)
Adding equation (i) and (ii), we get
2y - 6 = 0
⇒ 2y = 6
$\Rightarrow\ \text{y}=\frac{6}{2}=3$
Putting y = 3 in equation (i), we get
x + 3 - 3 = 0
⇒ x = 0
$\therefore$ (0, 3) be the coordinates of the point of intersection of the axis and directrix.
Then, coordinates of vertex (x₁, y₁) is the mid-point of the line segment joining (0, 3) and (-2, 1)
$\therefore\ \text{x}_1=\frac{0-2}{2}\text{ and }\text{y}_1=\frac{3+1}{2}$
⇒ required coordinates of vertex are (-1, 2).

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Question 111 Mark

PSQ is a focal chord of the parabola y² = 8x. If SP = 6, then write SQ.

Answer

The coordinates of the focal chord are $\text{P}(\text{at}^2, 2\text{at}^2)$ and $\text{Q}\Big(\frac{\text{a}}{\text{t}^2},\ \frac{-2\text{a}}{\text{t}}\Big).$
Comparing y² = 8x with y² = 4ax:
a = 2
Therefore, the coordinates of the focus S is (2, 0).
Given:
SP = 6
$\therefore\ \sqrt{(2-2\text{t}^2)^2+(4\text{t})^2=6}$
$\Rightarrow\ \text{t}^4+2\text{t}^2-8=0$
$\Rightarrow\ \text{t}^2=2$
Thus, we have:
$\text{SQ}=\sqrt{\Big(2-\frac{2}{\text{t}^2}\Big)^2+\Big(\frac{4}{\text{t}^2}\Big)}=\sqrt{\Big(2-\frac{2}{2}\Big)^2+\Big(\frac{4}{2}\Big)}=3$

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