Search Previous Year Questions

Calculation:

Given,

Line 1: $m^2x + ny - 1 = 0$

Line 2: $n^2x - my + 2 = 0$

Slope of Line 1, $m_1 = -\dfrac{m^2}{n}$

Slope of Line 2, $m_2 = \dfrac{n^2}{m}$

For perpendicular lines, $m_1 \times m_2 = -1$.

$\frac{-m^2}{n} \times \frac{-n^2}{m} = -1$

mn -1 = 0

Hence, the correct answer is Option 1.

Calculation:

Given points A(0,0), B(p,1), and C(1,q), which form an equilateral triangle. We are told (0 < p,q < 1).

Compute the squared lengths of the sides:

$AB^{2} = (p - 0)^{2} + (1 - 0)^{2} = p^{2} + 1$

$AC^{2} = (1 - 0)^{2} + (q - 0)^{2} = 1 + q^{2}$

$BC^{2} = (1 - p)^{2} + (,q - 1,)^{2} = (1 - p)^{2} + (q - 1)^{2} = 2,(1 - p)^{2}$

Because the triangle is equilateral, all three squared lengths are equal:

⇒ $AB^{2} = AC^{2}$

⇒ $p^{2} + 1 ;=; 1 + q^{2} ;\Longrightarrow; p^{2} = q^{2} ;\Longrightarrow; p = q$

both p and q are positive in (0,1)

Let p = q = t Then

⇒ $AB^{2} = t^{2} + 1$

⇒ $BC^{2} = 2,(1 - t)^{2}$

Equate AB² and BC²:

⇒ $t^{2} + 1 ;=; 2,(1 - t)^{2} ;=; 2\bigl(1 - 2t + t^{2}\bigr) ;=; 2 - 4t + 2t^{2}.$

Simplify:

⇒ $t^{2} + 1 = 2 - 4t + 2t^{2};\Longrightarrow;0 = 2 - 4t + 2t^{2} - (t^{2} + 1) = t^{2} - 4t + 1.$

So t satisfies:

⇒ $t^{2} - 4t + 1 = 0 \quad\Longrightarrow\quad t = \frac{4 \pm √{16 - 4}}{2} = \frac{4 \pm 2√{3}}{2} = 2 \pm √{3}.$

Since 0 < t < 1, we take t = 2 - √3 (note 2 + √3> 1) is not allowed. Hence,

⇒ $p = q = 2 - √{3}.$

Therefore:

 $p + q ;=; (2 - √{3}) + (2 - √{3}) ;=; 4 - 2√{3}.$

Hence, the correct answer is Option 4.

Calculation:

Given points A(1,1), B(0,0), and C(2,0). The angle bisectors meet at P (the incenter).

Compute the side lengths:

$a = BC = \sqrt{(2 - 0)^{2} + (0 - 0)^{2}} = 2$

$b = CA = \sqrt{(1 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}$

$c = AB = \sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}$

Therefore, $a + b + c = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}.$

By the incenter formula,

$X_{P} ;=; \frac{a,x_{A} + b,x_{B} + c,x_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 2,}{,2 + 2\sqrt{2},} ;=; \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} ;=; 1.$

$Y_{P} ;=; \frac{a,y_{A} + b,y_{B} + c,y_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 0,}{,2 + 2\sqrt{2},} ;=; \frac{2}{2 + 2\sqrt{2}} ;=; \frac{1}{,1 + \sqrt{2},} ;=; \sqrt{2} ;-; 1.$

∴ The incenter is $P = \bigl(1,;\sqrt{2} - 1\bigr)$.

Hence, the correct answer is Option 1.

Calculation:

Given points A(3, −1) and B(1, 1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P.

Compute the midpoint P of AB:

$P = \Bigl(\tfrac{3 + 1}{2},,\tfrac{-1 + 1}{2}\Bigr) = (2,,0)$

Compute the slope of AB:

$m_{AB} = \frac{,1 - (-1),}{,1 - 3,} = \frac{2}{-2} = -1$

Therefore, the equation of line AB is:

$y - 1 = -1,(x - 1);\Longrightarrow;x + y - 2 = 0$

The perpendicular bisector of AB must pass through P(2, 0) and have slope perpendicular to −1 (i.e. slope +1):

$y - 0 = 1,(x - 2);\Longrightarrow;y = x - 2$

Any point Q on this bisector satisfies y = x - 2. Write Q = (x, x − 2).

We require the distance PQ = √2. Since P(2, 0),

$\text{Distance}^2 = (x - 2)^2 + \bigl((,x - 2,) - 0\bigr)^2 = 2$

⇒ $(x - 2)^2 + (x - 2)^2 = 2;\Longrightarrow;2,(x - 2)^2 = 2;\Longrightarrow;(x - 2)^2 = 1$

Thus,

⇒ $x - 2 = \pm 1;\Longrightarrow;x = 3\text{ or }x = 1$

If x = 3, then y = 3 - 2 = 1. So one solution is Q(3, 1).

If x = 1, then y = 1 - 2 = -1. So the other solution is Q(1, −1).

Hence, the correct answer is Option 2.

Calculation:

Given vertices A(1, 2) and D(−2, 6), where AD is the altitude from A onto BC in an equilateral triangle ABC.

Compute the slope of AD:

$m_{AD} = \frac{,y_D - y_A,}{,x_D - x_A,} = \frac{,6 - 2,}{,-2 - 1,} = \frac{4}{-3} = -\tfrac{4}{3}.$

Because AD ⟂ BC, the slope of BC, $m_{BC}$, satisfies

$m_{AD} \cdot m_{BC} = -1 ;\Longrightarrow; \Bigl(-\tfrac{4}{3}\Bigr) ,m_{BC} = -1 ;\Longrightarrow; m_{BC} = \frac{-1}{, -\tfrac{4}{3},} = \tfrac{3}{4}.$

Equation of the line BC 

$y - 6 = \tfrac{3}{4},(x + 2).$

$4(y - 6) = 3(x + 2);\Longrightarrow;4y - 24 = 3x + 6.$

$4y - 24 - 3x - 6 = 0$

$;\Longrightarrow; -3x + 4y - 30 = 0 ;\Longrightarrow; 3x - 4y + 30 = 0.$

Hence, the correct answer is Option 4.

Calculation:

Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.

We have, 

$x - y = 0 ;\Longrightarrow; x = y.$

Substitute into x + y = 0:

$x + x = 0 ;\Longrightarrow; 2x = 0 ;\Longrightarrow; x = 0,; y = 0.$

Thus, the center is (0, 0).

Also radius,

$r = \frac{10}{2} = 5.$

the equation of the circle with center (0,0) and radius 5 is

$x^{2} + y^{2} = 5^{2} = 25.$

Hence, the correct answer is Option 2.

Calculation:

Given,

The circle’s equation is

$x^{2} + y^{2} + 2x + 2y + 1 = 0$

Rewrite by completing squares:

$(x + 1)^{2} + (y + 1)^{2} = 1$

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

$a\sqrt{2} = 2 ;\Longrightarrow; a = \sqrt{2}.$

Each vertex lies a half‐side $= \tfrac{a}{2} = \tfrac{1}{\sqrt{2}}$ from the center along both axes. Since the center is ( -1, -1), the four vertices are

$\displaystyle \bigl(-1 \pm \tfrac{1}{\sqrt{2}},; -1 \pm \tfrac{1}{\sqrt{2}}\bigr).$

One of these vertices is

$\bigl(-1 + \tfrac{1}{\sqrt{2}},; -1 - \tfrac{1}{\sqrt{2}}\bigr).$

Hence, the correct answer is Option 3

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

$m ;=; \tan(45^\circ) ;=; 1.$

A standard parametric form for y2 = 4x is 

$\bigl(x(t),,y(t)\bigr) ;=;\bigl(t^{2},,2t\bigr),$

since  $y^{2} = 4x \implies (2t)^{2} = 4,t^{2}$

The slope of the tangent at $\bigl(t^{2},,2t\bigr)$

Differentiate y2 = 4x 

$2y,\frac{dy}{dx} ;=; 4$

$;\Longrightarrow; \frac{dy}{dx} ;=; \frac{4}{,2y,} ;=; \frac{2}{,y,}.$

At the point $\bigl(x,y\bigr) = \bigl(t^{2},,2t\bigr)$ one has $y = 2t$

$\left.\frac{dy}{dx}\right|_{,(t^{2},,2t)}$ $= \frac{2}{,2t,} = \frac{1}{t}.$

We require this slope to equal 1.

$\frac{1}{t} ;=; 1 ;\Longrightarrow; t = 1.$

Now point of contact

Substitute t = 1 in $\bigl(x(t),,y(t)\bigr) = \bigl(t^{2},,2t\bigr)$. 

$x(1) = 1^{2} = 1,$

$y(1) = 2 \cdot 1 = 2.$

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

Calculation:

Given,

Hyperbola equation: $25x^{2} - 75y^{2} = 225$

Divide both sides by 225 to obtain standard form:

$\frac{25x^{2}}{225} - \frac{75y^{2}}{225} = 1 ;\Longrightarrow; \frac{x^{2}}{9} - \frac{y^{2}}{3} = 1$

Thus, $a^{2} = 9$ and $b^{2} = 3$.

Compute $c$ from $c^{2} = a^{2} + b^{2}$:

$c^{2} = 9 + 3 = 12 ;\Longrightarrow; c = \sqrt{12} = 2\sqrt{3}.$

The foci are at $(\pm c,,0)$, so the distance between them is $2c$:

$2c = 2 \times 2\sqrt{3} = 4\sqrt{3}.$

∴ The distance between the two foci is $4\sqrt{3}$ units.

Hence, the correct answer is Option 2.

Calculation:

Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,

$x = a,\sinα,\quad y = b,\cosα$

we identify

$a = 3,\quad b = 5$

Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is

$e = \sqrt{,1 - \frac{(\text{semi-minor})^{2}}{(\text{semi-major})^{2}},} = \sqrt{,1 - \frac{a^{2}}{b^{2}},}$

Substitute (a = 3) and (b = 5):

$e = \sqrt{,1 - \frac{3^{2}}{5^{2}},} = \sqrt{,1 - \frac{9}{25},} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

Hence, the correct answer is Option 2.