Quantitative Aptitude PYQs

Concept:

The directrix of the parabola y2 = - 4ax, having an x-axis, passes

through (a, 0), and has the equation x - a = 0.

Calculation:

Equation of parabola y2 = 4bx, where b < 0

b2 + b - 2 = 0

Solving: b2 + b - 2 = 0

⇒ b2 + 2b - b - 2 = 0

⇒ b (b + 2) - 1(b + 2) = 0

⇒ (b + 2) (b - 1) = 0

⇒ b = - 2, 1

As it is given that b < 0,

⇒ b = - 2

Equation of parabola, y2 = - 8x

Equation of directrix, x - 2 = 0.

Concept:

• The slope m of the line joining the two points (x₁, y₁) and (x₂, y₂) is
• m = \({(y_2-y_1) \over (x_2-x_1)}\)
• The slopes of parallel lines are the same.
• If the lines have the same slope and also have a common point, then they are the same line.

 

Calculation:

Let the points are A(-a, -b), B(0, 0), C(a, b) and D(a², ab).

Using the slope formula, the slope of the line AB will be:

m_AB = \({(-b-0) \over (-a-0)} = {b \over a}\)

Using the slope formula, the slope of the line BC will be:

mBC = \({(0-b) \over (0-a)} = {b \over a}\)

Using the slope formula, the slope of the line CD will be:

mCD = \({(b-ab) \over (a-a^2)} = {b \over a}\)

Using the slope formula, the slope of the line AD will be:

mAD = \({(-b-ab) \over (-a-a^2)} = {b \over a}\)

Since the slopes of all the lines are equal. Also, 

At a = 1, b = 1, points C & D will coincide. 

∴ The correct answer is option (D).

Alternate Method At a = 1, b = 1, the given point become 

A(-1, -1), B(0, 0), C(1, 1), D(1, 1)

Calculation:

Given that

16p2 + 49q2 - 4r2 - 56pq = 0

⇒ (4p)² + (7q)2 - (2r)² - 2 × 4p × 7q = 0

⇒ (4p)2 + (7q)2 - 2 × 4p × 7q - (2r)2 = 0

We know that, (a - b)² = a² + b² - 2ab

⇒ (4p - 7q)2 - (2r)2 = 0

⇒ (4p - 7q + 2r) (4p - 7q - 2r) = 0

⇒ 2(2p - 7/2q + r) 2(2p - 7/2q - r) = 0

⇒ (2p - 7/2q + r) (2p - 7/2q - r) = 0

On comparing with the (px + qy + r) (px + qy - r) = 0, we will get

∴ x = 2, y = -7/2.

Given: 

Equation of a chord, 3x + y - 5 = 0

Equation of the circle, x2 + y2 - 25 = 0

Formula Used:

General equation of circle of center (0, 0) & radius R is,

x2 + y2 = R²

Distance of a point (x₁,y₁) to a line Ax + By + C = 0 is

\(|{{Ax_1 + By_1 + C}\over \sqrt{A^2+B^2}}|\)

Distance between (x1,y1) and (x2,y2)

\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

 

Calculation:

The centre of the Circle is (0, 0) and Radius = 5,

Let the coordinates of the mid-point of the chord be (a, b)

(a, b) passes through 3x + y - 5 = 0,

3a + b = 5              ------(1)

Distance b/w (0, 0) to (a,b) = Distance b/w (0, 0) to (3x + y - 5 = 0)

\(\sqrt{a^2+b^2} = {{5}\over \sqrt{10}}\)

Squaring both sides,

\({a^2+b^2} = {{25}\over {10}}= \frac{5}{2}\)

Now, let's check from the option,

(a, b) is equal to \( \left(\frac{3}{2}, \frac{1}{2}\right) \)

Given: \(\frac{x^2}{24-k}+\frac{y^2}{k-16}=2\)

Concept:

Equation of an Ellipse is given by,

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Equation of a Hyperbola is given by,

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

Equation of a circle is given by,

\({x^2}+\ y^2=a^2\)

Calculation:

k = 19

\(\frac{x^2}{24-19}+\frac{y^2}{19-16}=2\)

\(\frac{x^2}{5}+\frac{y^2}{3}=2\)

Hence, the given equation is an ellipse if k = 19.

k = 12

\(\frac{x^2}{24-12}+\frac{y^2}{12-16}=2\)

\(\frac{x^2}{12}+\frac{y^2}{-4}=2\)

\(\frac{x^2}{12}-\frac{y^2}{4}=2\)

Hence, the given equation is hyperbola if k = 12.

k = 20

\(\frac{x^2}{24-20}+\frac{y^2}{20-16}=2\)

\(\frac{x^2}{4}+\frac{y^2}{4}=2\)

\({x^2}+{y^2}=8\)

Hence, the given equation is a circle if k = 20.

Concept:

The general equation of a hyperbola is given by,

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

where the X-axis is the transverse axis and the Y-axis is the conjugate axis. The foci are given by (ae, 0) and (-ae, 0)

The formula for the eccentricity of the hyperbola is \(e = \sqrt{(1 + ({b^2\over a^2})}\)

Calculation:

Equation of Hyperbola: 

\(\frac{x^2}{\cos ^2 \theta}-\frac{y^2}{\sin ^2 \theta}=1\)

Here, a² = cos²θ, b² = sin2θ

c² = a2 + b2 

c2 = cos2θ + sin2θ

c = 1

By using the above concept

\(e = \sqrt{(1 + ({b^2\over a^2})}\)

\(e = \sqrt{(1 + ({sin^2 \theta\over cos^2\theta})}\)

\(e = \sqrt{{cos^2\theta + {sin^2 \theta}\over cos^2\theta}}\)

e = 1/cos θ = sec θ

Hence, The eccentricity is sec θ.​

The foci are given by (ae, 0) and (-ae, 0)

ae = cos θ × sec θ = 1

The foci are given by (1, 0) and (-1, 0)

Hence, the two foci are independent of θ.

Distance between foci = 2

Hence, the distance between the two foci is 2 units. 

∴ All statements are correct.

Given:

Equation of a circle, 4x2 + 4y2 - 4ax - 4ay + a2 = 0

Concept:

The general equation of a circle is given by:

(x - h)² + (y - k)² = R² or

x2 + y2 - 2hx - 2ky + h2 + k2 - R2 = 0    

Where, (h, k) is the centre of the circle; R is the radius of the circle.

Calculation:

x² - 2hx + h² + y² - 2ky + y² = R²

Equation of a circle,

4x2 + 4y2 - 4ax - 4ay + a2 = 0

\(x^2 + y^2 - ax - ay + {a^2\over 4} = 0\)

By comparing:

-2h = -a, h = a/2

-2k = -a, k = a/2

So, center = (a/2, a/2)

h2 + k2 - R2 = a²/4

(a/2)² + (a/2)2 - R² = a²/4

⇒ R² = a2/4

⇒ R = a/2

⇒ D = 2R = a

The circle touches both the axes. 

 

The diameter of the circle is a.

The centre of the circle lies on the line x + y = a

Putting x, y as a/2, a/2, satisfies the equation x + y = a.

Hence, The centre of the circle lies on the line x + y = a

Hence, statements 1 and 3 are correct.

Concept:

If two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel then 

a₁/a₁ = b₁/b₂

Calculation:

The given equation of the linea are

(k - 3)x - (5 - k2) y + k2 - 7k + 6 = 0         -----(1)

x + y - 1 = 0                                            -----(2)

a₁ = k - 3, b₁ = -(5 - k²), a₂ = 1, b₂ = 1

By using the above concept, 

(k - 3)/1 = (k² - 5)/1

⇒ k - 3 = k² - 5

⇒ k² - k - 2 = 0

⇒ k2 - 2k + k - 2 = 0

⇒ k(k - 2) + 1(k - 2) = 0

⇒ (k - 2)(k + 1) = 0

⇒ k = 2, -1

∴ Correct answer is option (B).

Concept:

Internal Section Formula: 

Let A and B be the given two points (x1, y1) and (x2, y2) respectively and

P(x, y) be the point dividing the line segment AB internally in the ratio

m : n then 

⇔ P (x, y) = \(\rm (\dfrac{m_1x_2+m_2x_1}{m_1+m_2}, \dfrac{m_1y_2+m_2y_1}{m_1+m_2})\)

Calculation:

By using the above concept

x = \(\frac{k \times 5 + 1 \times (-1)}{k+1} = \frac{5k -1}{k+1}\)

y = \(\frac{k \times 7 + 1 \times 1}{k+1} = \frac{7k +1}{k+1}\)

Since, R(x, y) also lies on the line x + y = 4. Therefore,

\(\frac{5k - 1}{k+1} + \frac{7k +1}{k+1} = 4\)

\(⇒ \frac{5k - 1+7k+1}{k+1} = 4\)

⇒ 12k = 4(k + 1)

⇒ 12k - 4k = 4

⇒ 8k = 4

⇒ k = 4/8 = 1/2

Hence, the required ratio is PR ∶ RQ = 1 : 2

Concept:

If the normal to the line makes an angle θ with the positive x-axis and p

is the perpendicular distance from the origin to the line then, the

equation of the line is given by

x cos θ + y cos θ = p

\(cos:15° = {{\sqrt{3}+1}\over {2\sqrt2}}\), \(sin:15° = {{\sqrt{3}-1}\over {2\sqrt2}}\)

Calculation:

Given that,

The perpendicular distance from the origin = 4 units

The angle b/w normal to the positive direction of the x-axis = 15°

The equation of the line is given by

x cos 15° + y cos 15° = 4

Now intercept form is,

\({x\over {4\over cos:15°}} + {y\over {4\over sin:15°}} = 1\)

x-intercept = \(4\over cos:15°\),

y-intercept = \(4\over sin:15°\)

The sum of the intercept

S = \({4\over cos:15°} + {4\over sin:15°}\)

⇒ S = \(4({{{2\sqrt2}\over {\sqrt{3}+1}}} + {{{2\sqrt2}\over {\sqrt{3}-1}}})\)

⇒ S = \({8\sqrt2}({{{1}\over {\sqrt{3}+1}}} + {{{1}\over {\sqrt{3}-1}}})\)

⇒ S = \({8\sqrt2}({{{{\sqrt{3}-1}+{\sqrt{3}+1}}\over {3-1}}})\)

⇒ S = \({8\sqrt2}({{{{2\sqrt{3}}}\over {2}}})\)

∴ The sum of the intercept = \({8\sqrt6}\)

Given:

 \(\vec a=\rm \hat{i}+2 \hat{j}+3 \hat{k}\)  and \(\vec b =\rm2 \hat{i}+3 \hat{j}-2 \hat{k}\)

Formula Used:

The length of the projection of the vector \(\vec a\) on \(\vec b\) = \({{\vec a}.{\vec b}\over {{|\vec b|}}}\)

Calculation:

⇒ \({{​​(\rm \hat{i}+2 \hat{j}+3 \hat{k})}.{(\rm2 \hat{i}+3 \hat{j}-2 \hat{k})}\over {{\sqrt{2^2+3^2+2^2}}}}\)

⇒ \({{{​​2+6-6}}\over {{\sqrt{4+9+4}}}}\)

⇒ \({{{​​2}}\over {{\sqrt{17}}}}\)

The correct answer is option (2).

Formula Used:

\(\rm \vec{a} \times \vec{b}= |\vec{a}||\vec{b}| sin:\theta\)

\(\rm \vec{a} . \vec{b}= |\vec{a}||\vec{b}| cos:\theta\)

\(sin^2:\theta +cos^2:\theta=1\)

Given:  

\(\rm (\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=144\)

\(\rm|\vec{b}|=4 \)

Calculation:

\(\rm \vec{a} \times \vec{b}= |\vec{a}||\vec{b}| sin:\theta\)

\(\rm \vec{a} . \vec{b}= |\vec{a}||\vec{b}| cos:\theta\)

\(\rm (\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=144\)

\(\rm (|\vec{a}||\vec{b}| sin:\theta)^2+( |\vec{a}||\vec{b}| cos:\theta)^2=144\)

\(\rm |\vec{a}|^2|\vec{b}|^2 (sin^2:\theta +cos^2:\theta)=144\)

As \(sin^2:\theta +cos^2:\theta=1\)

\(\rm |\vec{a}|^2|\vec{b}|^2 =144\)

\(\rm |\vec{a}|^2={144 \over 16}\)

\(\rm |\vec{a}|={12\over 4}\)

\(\rm |\vec{a}|=3\)

Concept: 

\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos:\theta\)

Calculation:

Given that,  \(\vec{a} \cdot \vec{b} \geq 0\)

By using the above formula

\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos:\theta\)

As, \(\vec{a} \cdot \vec{b} \geq 0\)

\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|cos:\theta \geq 0\)

\(|\vec{a}||\vec{b}| \geq 0\) , \(cos:\theta \geq 0\)

\(cos:\theta\) is positive between 0 to π/2

∴ \(0 \leq \theta \leq \frac{\pi}{2}\)

Concept: 

If vectors \(\rm a_1 \hat{i}+b_1 \hat{j}\) and \(\rm a_2 \hat{i}+b_2 \hat{j}\) are collinear,

\({a_1 \over a_2} = {b_1\over b_2}\)

Calculation:

Given vectors are

\(\vec{A} = \rm 60 \hat{i}+3 \hat{j}, \vec{B} = 40 \hat{i}-8 \hat{j}\) and \(\vec{C} = \rm \beta \hat{i}-52 \hat{j}\)

Let the origin be: \(\rm 60 \hat{i}+3 \hat{j}\)

So, two vectors about the origin are

\(\vec{BA} = \rm (60-40) \hat{i}+(-8-3) \hat{j} \vec{BA} = \rm 20 \hat{i}-11 \hat{j}\)

\( \vec{CA} = \rm \beta \hat{i}-52 \hat{j} -\rm 60 \hat{i}-3 \hat{j}\vec{CA} =\rm ( \beta-60) \hat{i}+(-52-3) \hat{j} \vec{CA}= \rm (60-\rm \beta \hat{i}) \hat{i}-55 \hat{j}\)

Now, if vectors \(\rm 20 \hat{i}-11 \hat{j}\) and \(\rm (60-\rm \beta \hat{i}) \hat{i}-55 \hat{j}\) are collinear,

\({20 \over (60-\rm \beta)} = {-11\over -55}\)

\({20 \over (60-\rm \beta)} = {1\over 5}\)

\(100 = 60-\rm \beta\)

\(\rm \beta = -40\)

Given: 

\(\rm \vec{a}=(0,1,1)\) and \(\rm \vec{b}=(1,0,1) \)

Formula Used:

The angle between the vectors is given by

\(cos: \theta = {{\vec a.\vec b}\over {|\vec a|.|\vec b|}}\)

Unit Vectors = \(\pm {{\vec{a} \times\vec{b}}\over |{\vec{a} \times\vec{b}}|}\)

Calculation:

\(\rm \vec{a}\) = j + k

\(\rm \vec{b}\) = i + k

Unit Vectors = \(\pm {{\vec{a} \times\vec{b}}\over |{\vec{a} \times\vec{b}}|}\)

Hence, statement I is incorrect.

\(cos: \theta = {{\vec a.\vec b}\over {|\vec a|.|\vec b|}}\)

\(cos: \theta = {{0+0+1}\over {{\sqrt{0^2+1^2+1^2}.{\sqrt{1^2+0^2+1^2}}}}}\)

\(cos: \theta = {{1}\over {{\sqrt{2}.{\sqrt{2}}}}}\)

\(cos: \theta = {{1}\over {2}}\)

\( \theta = {{\pi}\over {3}}\)

Hence, statement II is correct.

∴ The angle between the vectors is \(\frac{\pi}{3}\).

Given:

L is the line with direction ratios < 3, -2, 6 >, passing through (1, -1, 1)

Formula Used:

Equation of a line which passed through (x1,y1,z1) is given by

\(\rm \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\)

Where, l, m, n are the Direction Ratios

Distance of (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by

\({\sqrt {{(x_2-x_1)}^2+{(y_2-y_1)^2}+{(z_2-z_1)^2}}}\)

Calculation:

Equation of L,

\(\rm \frac{x-1}{3}=\frac{y+1}{-2}=\frac{z-1}{6} = k\) (let)

(x, y, z) = (3k + 1, -2k - 1, 6k + 1)

Distance of (3k + 1, -2k - 1, 6k + 1) & (1, -1, 1) is 2 units

⇒ \({\sqrt {{(3k+1-1)}^2+{(2k - 1+1)^2}+{(6k + 1 - 1)^2}}} = 2\)

⇒\({\sqrt {{(3k)}^2+{(2k )^2}+{(6k)^2}}} = 2\)

⇒\({\sqrt {{9k}^2+{4k^2}+{36k^2}}} = 2\)

⇒ \(7k = \pm2\)

⇒ \(k = \pm{2 \over7}\)

Putting the value of k in (3k + 1, -2k - 1, 6k + 1) we get, 

\(\left(\frac{13}{7},-\frac{11}{7}, \frac{19}{7}\right)\) and \(\left(\frac{1}{7},-\frac{3}{7},-\frac{5}{7}\right)\)

Concept:

If a plane is parallel to a line then, the sum of the product of direction ratios of the line and the plane is zero.

Calculation:

Given equation of a Line is

\(\rm \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)

Direction Ratio's of line = 3, 4, 5.

By checking the options,

Option 1: Dr's of 2x + 2y + z - 1 = 0 are 2, 2, 1

Product of direction ratios = 3×2 + 4×2 + 5×1 = 6 + 8 + 5 = 19

Option 2: Dr's of 2x - y - 2z + 5 = 0 are 2, -1, -2

Product of direction ratios = 3×2 + 4×-1 + 5×-2 = 6 - 4 - 10 = -8

Option 3: Dr's of 2x + 2y - 2z + 1 = 0 are 2, 2, -2

Product of direction ratios = 3×2 + 4×2 + 5×-2 = 6 + 8 - 10 = 4

Option 4: Dr's of x - 2y + z - 1 = 0 are 1, -2, 1

Product of direction ratios = 3 × 1+ 4 × -2 + 5 × 1 = 3 - 8 + 5 = 0

Hence, option 4 is correct.

Concept:

 

The angle between these lines, \({{x-x_1} \over a_1} = {{y-y_1} \over b_1}= {{z-z_1} \over c_1}\) and \({{x-x_2} \over a_2} = {{y-y_2} \over b_2}= {{z-z_2} \over c_2}\) is given by,

\(cos:\theta = {{a_1.a_2}+{b_1.b_2}+{c_1.c_2}\over {{\sqrt{a^2_1 +b^2_1+c^2_1}}.{{\sqrt{a^2_2 +b^2_2+c^2_2}}}}}\)

Calculation:

Given:

2x = 3y = -z 

\({x \over 3} = {y \over 2} = {z \over -6}\)       ----(1)

6x = -y = -4z 

\({x \over 2} = {y \over -12} = {z \over -3}\)         ----(2)

By using the above concept for line (1) & (2)

\(cos:\theta = {{3.(2)}+{2.(-12)}+{(-6).(-3)}\over {{\sqrt{3^2 +2^2 +(-6)^2}}.{{\sqrt{2^2 +(-12)^2+(-3)^2}}}}}\)

\(cos:\theta = {{6-24+18}\over {{\sqrt{3^2 +2^2 +(-6)^2}}.{{\sqrt{2^2 +(-12)^2+(-3)^2}}}}}\)

\(cos:\theta = 0\)

∴ \(\theta = 90°\)

Concept:

Sphere: General equation of sphere is given as

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0

• Centre of sphere (-u, -v, -w)
• Radius, \({\rm{r}} = √{({{\rm{u}}^2} + {{\rm{v}}^2} + {{\rm{z}}^2} - {\rm{d}})} ​​\)

 

Equation of sphere with center (a, b, c) and radius r is given by

(x - a)2 + (y - b)2 + (z - c)2 = r2

Calculation:

Equation of given sphere is 

x2 + y2 + z2 - 2x - 6y - 8z - 5 = 0

Centero of sphere is (1, 3, 4)

Let the equation of the sphere concentric with given sphere be 

x2 + y2 + z2 - 2x - 6y -8z + k = 0

As the sphere passes through the origin, this must satisfy (0, 0).

x2 + y2 + z2 - 2x - 6y - 8z + k = 0

⇒ 02 + 02 + 02 - 2(0) - 6(10) - 8(0) + k = 0

⇒ k = 0

∴ Required equation is x2 + y2 + z2 - 2x - 6y - 8z = 0

Formula Used:

If a line joined by (x1, y1, z1) and (x2, y2, z2) is divided by (x, y, z) in the ratio k:1

\((x,y,z) = ({1\times x_2+ k \times x_1 \over 1+k},{1\times y_2+ k \times y_1 \over 1+k},{1\times z_2+ k \times z_1 \over 1+k})\)

Calculation:

Given that

A point P lies on the line joining A(1, 2, 3) and B(2, 10, 1)

Let (1, 2, 3) be (x1, y1, z₁) and (2, 10, 1) be (x2, y2, z₂) and  (x, y, 7) be (x, y, z)

So, (x, y, 7) bisect (1, 2, 3) and (2, 10, 1) in ratio k :1

\((x,y,z) = ({1\times x_2+ k \times x_1 \over 1+k},{1\times y_2+ k \times y_1 \over 1+k},{1\times z_2+ k \times z_1 \over 1+k})\)

\((x,y,7) = ({1\times 1+ k \times 2 \over 1+k},{1\times 2+ k \times 10 \over 1+k},{1\times 3+ k \times 1 \over 1+k})\)

By Equating z - coordinate, 

\({1\times 3+ k \times 1 \over 1+k} = 7\)

\(3 + k = 7 + 7k\)

k = - 4/6 = -2/3

x + y = \({1\times 1+ k \times 2 \over 1+k} + {1\times 2+ k \times 10 \over 1+k}\)

x + y = \({1+ 2k \over 1+k} + {2+ 10k\over 1+k}\)

x + y = \({1+ 2k +2+ 10k\over 1+k}\)

x + y = \({3+ 12k\over 1+k}\)

Put k = -2/3

∴ x + y = -15