MATHS M.C.Q (1 Marks)

3. $\Big(-\frac{1}{4},0\Big)$

Solution:

Given:

Equation of  the parabola = y = 2x² + x

$\Rightarrow\ \text{x}^2+\frac{\text{x}}{2}=\frac{\text{y}}{2}$

$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{\text{y}}{2}+\frac{1}{16}$

$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{8\text{y}+1}{16}$

$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{1}{2}(\text{y}+\frac{1}{8})$

$\text{Let }\text{X}=\text{x}+\frac{1}{4},\text{Y}=\text{y}+\frac{1}{8}$

$\therefore\ \text{X}^2=\frac{1}{2}\text{Y}$

Comparing with X = 4aY

$\text{a}=\frac{1}{8}$

Focus $=(\text{X}=0,\ \text{Y}=\text{a})=\Big(\text{x}=\frac{-1}{4},\text{y}=0\Big)$

Hence, the focus is at $\Big(-\frac{1}{4},0\Big).$

1. 3x + 2y + 14 = 0

Solution:

Given:

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

$\therefore$ 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).

$\therefore\ \frac{\text{r+2}}{2}=-1,\ \frac{\text{s}+3}{2}=1$

$\Rightarrow\ \text{r}=-4,\ \text{s}=-1$

Equation of the directrix:

$(\text{y}+1)=\frac{-3}{2}(\text{x}+4)$

$\Rightarrow\ 3\text{x}+2\text{y}+14=0$

3. (-1, 2)

Solution:

Let the coordinates of P and Q be (at₁², 2at₁) and (at₂², 2at₂), respectively.

$\text{Slope of PQ }=\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}...(1)$

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

$\therefore\ \text{Slope of PQ}=\frac{-2}{-1}=2$

From (1),

$\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}=2\ ...(2)$

Putting 4a = 8,

a = 2

$\therefore$ Focus of the given parabola = (a, 0) = (2, 0)

Using equation (2):

$\frac{4(\text{t}_2-\text{t}_1)}{2(\text{t}_2^2-\text{t}_1^2)}=2$

$\frac{(\text{t}_2-\text{t}_2)}{(\text{t}_2^2-\text{t}_1^2)}=1$

$\Rightarrow\ \text{t}_1+\text{t}_2=1$

As, points P and Q lie on 2x - y + 4 = 0

⇒ P(at₁², 2at₁) or P(2t₁², 4t₁) lie on line 2x - y + 4 = 0

⇒ 2(2t₁²) - (4t₁) + 4 = 0

⇒ t₁² - t₁ + 1 = 0 ...(3)

Also, Q(at₂², 2at₂) or P(2t₂², 4t₂) lie on line 2x - y + 4 = 0

⇒ 2(2t₂²) - (4t₂) + 4 = 0

⇒ t₂² - t₂ + 1 = 0 ...(4)

Adding (3) and (4), we get,

⇒ t₁² - t₁ + 1 + t₂² - t₂ + 1 = 0

⇒ (t₁² + t₂²) - (t₁ + t₂) + 2 = 0

⇒ (t₁² + t₂²) - 1 + 2 = 0 [t₁ + t₂ = 1, proved above]

⇒ (t₁² + t₂²) = -1

Let (x₁, y₁) be the mid-point of PQ.

Then, we have:

$\text{y}_1=\frac{2\text{at}_2+2\text{at}_1}{2}=2(\text{t}_1+\text{t}_2)=2$

And, $\text{x}_1=\frac{\text{at}_1^2+\text{at}_2^2}{2}=\text{t}_1^2+\text{t}_2^2=-1$

$\Rightarrow\ (\text{x}_1, \text{y}_1)=(-1, 2)$

2. x² - 2xy + y² + 6ax + 10ay - 7a² = 0 

Solution:

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.

$\therefore$ Equation of the axis of the parabola = y − 0 = 1(x - a) ...(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 (2a, a)

The vertex is the mid-point of the segment joining K and the focus (h, k).

$\therefore\ \text{a}=\frac{2\text{a+h}}{2},\ 0=\frac{\text{a+k}}{2}$

h = 0, k = -a

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:

SP = PM

⇒ SP² = PM²

^{$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$}

$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$

$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$

$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$

2. $\frac{1}{2}$

Solution:

Given:

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

The given equation of parabola can be rewritten as follows:

(y + 3)² - 9 + 5 + 2x = 0

⇒ (y + 3)² + 2x = 4

⇒ (y + 3)² = 4 - 2x

⇒ (y + 3)² = -2(x - 2)

Let Y = y + 3, X = x - 2

Then, the equation of parabola becomes Y² = -2X.

Vertex = (X = 0, Y = 0) = (x - 2 = 0, y + 3 = 0) = (x = 2, y = -3)

Comparing with y² = 4ax:

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

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

$\Rightarrow\ \text{SV}=\sqrt{\Big(2-\frac{3}{2}\Big)^2+(-3+3)^2}=\frac{1}{2}$

3. $ (−1, 2)$

Solution:

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.

$\therefore$ Equation of the axis of the parabola = y - 1 = 1(x + 2) ...(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.

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

h = -1, k = 2

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

1. (1, 2)

Solution:

Given:

(y - 2)² = 16 (x - 1)

Let X = x - 1, Y = y - 2

$\therefore$ Y² = 16X

Vertex = (X = 0, Y = 0) = (x - 1 = 0, y - 2 = 0) = (x = 1, y = 2)

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

1. $4\sqrt2\text{a}$

Solution:

Let OP be the chord.

Let the coordinates of P be (x₁, y₁).

From the figure, we have:

OP² = x₁² + y₁² ...(1)

$\text{And },\tan\frac{\pi}{4}=\frac{\text{y}_1}{\text{x}_1}$

⇒ x₁ = y_{1 ...(2)}

Also, (x₁, y₁) lies on the parabola.

$\therefore$ y₁² = 4ax_{1 ...(3)}

Using (2) and (3):

x₁² = 4ax₁ ⇒ x₁ = 4a ...(4)

From (4), (1) and (2), we have:

OP² = (4a)² + (4a)² = 32a²

$\Rightarrow\ \text{OP}=4\sqrt2\text{a}$

Therefore, the length of the chord is $4\sqrt2\text{a}$ a units.

1. x = 0

Solution:

Given:

x = t² + 1 ...(1)

y = 2t + 1 ...(2)

From (1) and (2):

$\text{x}=\Big(\frac{\text{y}-1}{2}\Big)^2+1$

On simplifying:

(y - 1)² = 4(x - 1)

Let Y = y - 1 and X = x - 1

$\therefore$ Y² = 4X

Comparing it with y² = 4ax:

a = 1

Therefore, the equation of the directrix is X = -a , i.e. x - 1= -1 ⇒ x = 0

1. x + 2y = 4

Solution:

Given:

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

$\therefore$ 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).

$\therefore\ \frac{\text{r}+2}{2}=1,\ \frac{\text{s}+6}{2}=4$

$\Rightarrow\ \text{r}=0,\ \text{s}=2$

Equation of the directrix:

$(\text{y}-2)=\frac{-1}{2}(\text{x}-0)$

⇒ x + 2y = 4

3. 8

Solution:

y² + 8x - 2y + 17 = 0

⇒ (y - 1)² - 1 + 8x + 17 = 0

⇒ (y - 1)² + 8x + 16 = 0

⇒ (y - 1)² = -8(x + 2)

Let X = x + 2, Y = y - 1

$\therefore$ Y² = -8X

Comparing with y² = 4ax:

a = 2

Length of the latus rectum = 4a = 8 units

3. 8

Solution:

Given:

x² - 4x - 8y + 12 = 0

(x - 2)² - 8y + 8 = 0

(x - 2)² = 8y - 8 = 8(y - 1)

Let X = x - 2, Y = y - 1

$\therefore$ X² = 8Y

$\therefore$ Length of the latus rectum = 4a = 8 units

3. Parabol

Solution:

Suppose PQ is a double ordinate of the parabola y² = 4ax.

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

Let (h, k) be the coordinates of R.

Then, we have:

OL = h  and RL = k

$\therefore$ RS = RL + LS = k + k = 2k

⇒ PR = RS = SQ = 2K

⇒ LP = LR + RP = k + 2k = 3k

Thus, the coordinates of P are (h, 3k) which lie on y² = 4ax.

$\therefore$ 9k² = 4ah

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

4. x² + y² - 2xy - 18x - 10y - 45 = 0

Solution:

Let P (x, y) be any point on the parabola whose focus is S (1, -1) and the directrix is x + y+ 7 = 0.

Draw PM perpendicular to x + y + 7 = 0.

Then, we have:

SP = PM

⇒ SP² = PM²

$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{1+1}}\Big)^2$

$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{2}}\Big)^2$

$\Rightarrow\ 2(​\text{x}^2+1-2\text{x}+\text{y}^2+1+2\text{y}​)\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$

$\Rightarrow\ (​2\text{x}^2+2-4\text{x}+2\text{y}^2+2+4\text{y}​)\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$

$\Rightarrow\ \text{x}^2+\text{y}^2-45-10\text{y}-2\text{xy}-18\text{x}=0$

Hence, the required equation is x² + y² - 2xy - 18x - 10y - 45 = 0.

1. x² + y² - 2xy + 8x + 8y - 16 = 0

Solution:

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:

SP = PM

⇒ SP² = PM²

^{$\Rightarrow\ (\text{x}-0)^2+(\text{y}-0)^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$}

^{$\Rightarrow\ \text{x}^2+\text{y}^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$}

⇒ 2x² + 2y² = x² + y² + 16 + 2xy - 8x - 8y

⇒ x² + y² - 2xy + 8x + 8y - 16 = 0

1. $\frac{1}{2}$

Solution:

Given:

4y² + 2x - 20y + 17 = 0

$\Rightarrow\ \text{y}^2+\frac{\text{x}}{2}-5\text{y}+\frac{17}{4}=0$

$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2+\frac{\text{x}}{2}-2=0$

$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=-1\Big(\frac{\text{x}}{2}-2\Big)$

$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=\frac{-1}{2}(\text{x}-4)$

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

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

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

1. $\Big(\frac{5}{4}, 1\Big)$

Solution:

Given:

The equation of the parabola is y² - x - 2y + 2 = 0.

⇒ (y - 1) - 1 = (x - 2)

(y - 1) = x - 1

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

Y = X

Comparing with Y = 4aX:

$\text{a}=\frac{1}{4}$

Focus=

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

Hence, the focus is at $\Big(\frac{5}{4}, 1\Big)$