MATHS M.C.Q (1 Marks)

2. 4 (x² + y² - x - y) + 1 = 0

Solution:

The line 4x + 3y = 6 cuts the coordinate axes at $\Big(\frac{3}{2},\ 0\Big)$ and (0, 2)

The coordinates of the incentre is $\Big(\frac{\text{ax}_1+\text{bx}_2+\text{cx}_3}{\text{a+b+c}},\ \frac{\text{ay}_1+\text{by}_2+\text{cy}}{\text{a+b+c}}\Big)$

Here, $\text{a}=\frac{5}{2},\ \text{b}=\frac{3}{2},\ \text{c}=2,\ \text{x}_1=0,\ \text{y}_1\\=0,\ \text{x}_2=0,\ \text{y}_2=2,\ \text{x}_3=\frac{3}{2},\ \text{y}_3=0$

Thus, the coordinates of the incentre:

$\Big(\frac{0+0+3}{6},\ \frac{0+3+0}{6}\Big)$

$=\big(\frac{1}{2},\ \frac{1}{2}\Big)$

The equation of the incircle:

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

Also, radius of the incircle $=\frac{\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}}{\text{s}}$

Here, $\text{s}=\frac{\text{a+b+c}}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$

$\therefore$ Radius of the incircle $=\sqrt{\frac{3(3-\text{a})(3-\text{b}(3-\text{c}))}{3}}$

$=\frac{\sqrt{3\Big(3-\frac{5}{2}\Big)\Big(3-\frac{3}{2}\Big)(3-\text{c})}}{3}$

$=\frac{\sqrt{3\Big(3-\frac{1}{2}\Big)\Big(\frac{3}{2}\Big)}}{3}$

$=\frac{1}{2}$

The equation of circle:

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

$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$

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

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

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

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

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

3. $(-\infty,\ -3)\cup(4,\ \infty)$

Solution:

The given equations of the circles are x² + y² + x - 2y − 14 = 0 and x² + y² = 13.

Since (2, k) lies outside the given circles, we have:

4 + k² + 2 - 2k - 14 > 0 and 4 + k² > 13

⇒ k² - 2k - 8 > 0 and k² > 9

⇒ (k - 4)(k + 2) > 0 and k² > 9

⇒ k > 4 or k < -2 and k > 3 or k < -3

⇒ k > 4 and k < -3

$\Rightarrow\text{k}\in(-\infty,\ -3)\cup(4,\ \infty)$

1. A point

Solution:

The radius of the given circle $=\sqrt{1^1+(-2)^2-5}=0$

Hence, the radius of the given circle is zero, which represents a point.

2. $\frac{3}{4}$

Solution:

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

for a circle of a $(\text{x}-\alpha)^2+6 (\text{y}-\beta)^2=1$ then (a = 6)

$\frac{\lambda}{3}=\frac{1}{4}$

$\lambda=\frac{3}{4}$

1. $\text{e}=\frac{\sqrt{5}-1}{2}$

Solution:

According to the question, the distance between the foci is equal to the length of the latus rectum.

$\frac{2\text{b}^2}{\text{a}}=2\text{ae}$

$\Rightarrow\text{b}^2=\text{a}^2\text{e}$

Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$

$\Rightarrow\text{e}=\sqrt{1-\frac{\text{a}^2\text{e}}{\text{a}^2}}$

$\Rightarrow\text{e}=\sqrt{1-\text{e}}$

On squaring both sides, we get:

$\text{e}^2\text{e}-1=0$

$\Rightarrow\text{e}=\frac{-1\pm\sqrt{1+4}}{2}$

$\Rightarrow\text{e}=\frac{\sqrt{5}-1}{2}$ $(\because\text{e}\text{ cannot be negative}\big)$ 

1. x² + y² - 2x - 2y - 3 = 0

Solution:

The centre of the circumcircle is (1, 1).

Radius of the circumcircle

$\therefore$ Equation of the circle: $=\sqrt{(1+1)^2+(1-2)^2}=\sqrt{5}$

(x - 1)² + (y - 1)² = 5

⇒ x² + y2 - 2x - 2y - 3 = 0

3. $\frac{2}{\sqrt{3}}$

Solution:

Let the equation of the hyperbola be $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

Length of latus rectum = 8

$\therefore\ \frac{2\text{b}^2}{2}=8$

$\Rightarrow\text{b}^2=4\text{a}$

Conjugate axis = half of the distance between the foci

$\therefore\ 2\text{b}=\text{ae}$

Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$

From eqs. (i) and (iii), we get

$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$

$\Rightarrow\text{e}^2=4\text{e}^2-4$

$\Rightarrow\text{e}^2=\frac{4}{3}$

$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$

4. 2ae²

Solution:

 Length of the latus rectum $=\frac{2\text{b}^2}{\text{a}}$

and $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$

$\text{a}^2\text{e}^2=\text{a}^2-\text{b}^2$

$\Rightarrow\text{b}^2=\text{a}^2-\text{a}^2\text{e}^2$

$\Rightarrow\text{b}^2=\text{a}^2(1-\text{e}^2)$

$\therefore\ $Length of the latus rectum $=\frac{2\text{a}^2(1-\text{e}^2)}{\text{a}}=2\text{a}(1-\text{e}^2)$

Length of the major axis = 2a

Difference between length of latus rectum and length of major axis $=2\text{a}-2\text{a}(1-\text{e}^2)$

$\\=2\text{a}-2\text{a}+2\text{ae}^2\\=2\text{ae}^2$

1. g² < c

Solution:

Given:

x² + y² + 2gx + 2fy + c = 0 ......... (1)

The given circle intersects the x-axis.

The equation of circle becomes x² + 2gx + c = 0 ......... (2)

Solving equation (2):

$\therefore$ Discriminant, $\text{D}=\sqrt{4\text{g}^2-4\text{c}}\geq0$

$\Rightarrow4\text{g}^2-4\text{c}\geq0$

$\Rightarrow\text{g}^2-\text{c}\geq0$

$\Rightarrow\text{g}^2\geq\text{c}$

Hence, if g² < c, then the given circle will not intersect the x-axis.

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

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

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).

3. 16

Solution:

$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$

$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$

$\lambda^2+(\lambda-1)^2\leq120$

$\Rightarrow2\lambda^2-2\lambda-199\leq0$

Using quadratic formula:

$\Rightarrow\lambda=\frac{2\pm\sqrt{2^2-4(2)(-119)}}{2(2)}$

$\Rightarrow\lambda=\frac{2\pm\sqrt{956}}{4}$

$\Rightarrow\lambda=\frac{1\pm\sqrt{239}}{2}$

$\Rightarrow\lambda=-7.23,\ 8.23$

$\Rightarrow-7.23\leq\lambda\leq8.23$

$\Rightarrow\lambda=-7,\ -6,\ -5,\ -4,\ -3,\ -2,\ -1\\0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ $ $(\text{if}\ \lambda\in\text{Z})$

Thus, the number of integral values of $\lambda$ is 16.

1. $\frac{225\sqrt{3}}{6}$

Solution:

Let ABC be the required equilateral triangle.

The equation of the circle is x² + y² - 6x - 8y - 25 = 0.

Therefore, coordinates of the centre O is (3, 4).

Radius of the circle $=\text{OA}=\text{OB}=\text{OC}=\sqrt{9+16+25}=5\sqrt{2}$

In $\Delta\text{BOD},$ we have:

$\sin60^\circ=\frac{\text{DB}}{\text{BO}}$

$\Rightarrow\text{DB}=\frac{\sqrt{3}}{2}(5\sqrt{2})$

$\Rightarrow\text{BC}=2\text{BD}-\sqrt{3}\big(5\sqrt{2}\big)=5\sqrt{6}$

Now, area of $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{BC}^2=\big(5\sqrt{6}\big)^2\\=\frac{\sqrt{3}(150)}{4}=\frac{\sqrt{3}(75)}{2}=\frac{\sqrt{3}(225)}{6}$ square units

1. $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$

Solution:

Given:

x² + y² + 2ax + c = 0 ....... (1)

And, x² + y² + 2by + c = 0 ........ (2)

For circle (1), we have:

Centre = (-a, 0) = C₁

For circle (2), we have:

Centre = (0,-b) = C₂

Let the circles intersect at point P.

$\therefore$ Coordinates of P = Mid point of C₁C₂

$\Rightarrow$ Coordinates of P $=\Big(\frac{-\text{a}+0}{2},\ \frac{0-\text{b}}{2}\Big)=\Big(\frac{-\text{a}}{2},\ \frac{-\text{b}}{2}\Big)$

Now, we have:

PC₁ = radius of (1)

$\Rightarrow\sqrt{(-\text{a}+\frac{\text{a}}{2})^2}+\Big(0-\frac{\text{b}}{2}\Big)^2=\sqrt{\text{a}^2-\text{c}}$

$\Rightarrow\frac{\text{a}^2}{4}+\frac{\text{b}}{4}^2=\text{a}^2-\text{c}\ .....(3)$

Also, radius of circle (1) = radius of circle (2)

$\Rightarrow\sqrt{\text{a}^2-\text{c}}=\sqrt{\text{b}^2-\text{c}}$

$\Rightarrow\text{a}^2=\text{b}^2\ .....(4)$

From (3) and (4), we have:

$\frac{\text{a}^2}{2}=\text{a}^2-\text{c}$

$\Rightarrow\frac{\text{a}^2}{2}=\text{c}$

$\Rightarrow\frac{2}{\text{a}^2}=\frac{1}{\text{c}}$

$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{a}^2}=\frac{1}{\text{c}}$

$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$

2. (2, 0)

Solution:

Given parabola equation y² = 8x …(1)

Here, the coefficient of x is positive and the standard form of parabola is y² = 4ax … (2)

Comparing (1) and (2), we get

4a = 8

$\text{a} = \frac{8}{4} = 2$

We know that the focus of parabolic equation y² = 4ax is (a, 0).

$\therefore$ The focus of the parabola y² = 8x is (2, 0).

2.  $\Big(\frac{2}{3},\ -1\Big)$

Solution:

To find the centre:

Coefficient of x² = Coefficient of y²

$\therefore\lambda=2\lambda-3\Rightarrow\lambda=3$

Therefore, the given equation can be rewritten as $3\text{x}^2+3\text{y}^2-4\text{x}+6\text{y}-1=0.$

$\therefore\text{x}^2+\text{y}^2-\frac{4}{3}\text{x}+2\text{y}-\frac{1}{3}=0$

Thus, the coordinates of the centre is $\Big(\frac{2}{3},\ -1\Big).$

1. $\big(3, \frac{3}{2}\big)$

Solution:

x² - 2xy - 3x + 6y = 0

⇒ (x - 3) (x - 2y) = 0

⇒ x = 3 and x = 2y are two normals.

The intersection point of these two normals will be the centre of the circle.

$\therefore$ for x = 3

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

The intersection point is $\big(3, \frac{3}{2}\big)$ the centre of the given circle is $\big(3, \frac{3}{2}\big)$

2.   a² = b²(1 - e²)

Solution:

Given equation is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b})$

$\therefore\text{ Eccentricity e}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}$

$\Rightarrow\text{e}^2=1-\frac{\text{a}^2}{\text{b}^2}$

$\Rightarrow\frac{\text{a}^2}{\text{b}^2}=(1-\text{e}^2)$

$\Rightarrow\text{a}^2=\text{b}^2(1-\text{e}^2)$  

1. 1

Solution:

x² + y² = a ........ (1)

And, x² + y² − 6x − 8y + 9 = 0 ........ (2)

Let circles (1) and (2) touch each other at point P.

The centre of the circle x² + y² = a, 0, is (0, 0).

The centre of the circle x² + y² − 6x − 8y + 9 = 0, C₁, is (3, 4).

Also, radius of circle (1) $=\sqrt{\text{a}}=\text{OP}$

Radius of circle (2) $\sqrt{9+16-9}=4=\text{C}_1\text{P}$

From figure, we have:

$\Rightarrow\sqrt{3^2+4^2}=4+\sqrt{\text{a}}$

$\Rightarrow5=4+\sqrt{\text{a}}$

$\Rightarrow\text{a}=1$

3. $(\pm5,0)$

Solution:

The equation of the hyperbola is given below:

9x² − 16y² = 144

This equation can be rewritten in the following way:

$\frac{9\text{x}^2}{144}-\frac{16\text{y}^2}{144}=1$

$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{9}=1$

This is the standard equation of a hyperbola, where a² = 16 and b² = 9.

The eccentricity is calculated in the following way:

b² = a²(e² − 1)

⇒ 9 = 16(e² − 1)

$\Rightarrow\frac{9}{16}=\text{e}^2-1$

$\Rightarrow\text{e}=\frac{5}{4}$

$\text{Foci}=(\pm\text{ae},0)=(\pm5,0)$

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

Solution:

 According to the question, the minor axis is equal to the distance between the foci.

i.e. $2\text{b}=2\text{ae}$

$\text{e}=\frac{\text{b}}{\text{a}}\ \dots(1)$

Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$

$\Rightarrow\text{e}=\sqrt{1-\text{e}^2}$ $\Big[\text{From}\ (1)\Big]$ 

On squaring both sides, we get:

$\text{e}^2=1-\text{e}^2$

$\Rightarrow2\text{e}^2=1$

$\Rightarrow\text{e}^2=\frac{1}{2}$

$\Rightarrow\text{e}=\frac{1}{\sqrt{2}}$

4. $\sqrt { 20 }$

Solution:

Point of intersection of the given diameters is (8, -2) which is the centre of the circle.

Also the circle pass through the point (6, 2) so the radius is.

$=\sqrt{ (8-6)^2+(-2-2)^2}=\sqrt{20}$

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.

2. (-1, 2)

Solution:

We have,

y² - 4y - x + 3 = 0

(y - 2)² - 4 - x + 3 = 0

(y - 2)² = (x + 1)

$\therefore$ Vertex of the parabola = (-1, 2)

1. A hyperbola.

Solution:

Let P(x, y) be any point on the hyperbola $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1.$

By definition, we have:

$\text{PA}=\text{e}\big(\text{x}-\frac{\text{a}}{\text{e}}\big)=\text{ex}-\text{a}$

$\text{and }\text{PB}=\text{e}\big(\text{x}+\frac{\text{a}}{\text{e}}\big)=\text{ex}+\text{a}$

$\therefore\text{PB}−\text{PA}=\text{(ex+a)}−\text{(ex}−\text{a})=\text{2a = k}$

2. $225\pi$

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

2. $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$

2. $\pm4$

Solution:

The equation of the circle is x² + y² + 2ax + 8y + 16 = 0.

Its centre is (-a, -4) and its radius is a units.

Since the circle touches the x-axis, we have:

$\sqrt{(-\text{a}+\text{a})^2+(4-0)^2}=\text{a}$

$\Rightarrow\text{a}=\pm4$

1. x² + y² - 4x + 2y = 45

Solution:

Equation of circle, $(\text{x} - 2)^2 + (\text{y} -( -1))^2= \Big(\sqrt{{(3-2)^2+(6}-(-1))^2\Big)}^2$

$\text{x}^2 - 4\text{x} + 4 + \text{y}^2 + 2\text{y} + 1=(\sqrt{1+49})^2\therefore\text{x}^2+\text{y}^2-4\text{x}+2\text{y}=45$ Equation of circle.

3. $\sqrt{\frac{3}{2}}$

Solution:

The lengths of the latus rectum and the transverse axis are $\frac{2\text{b}^2}{\text{a}}\text{ and }2\text{a},$ respectively.

According to the given statement, length of the latus rectum is half of its transverse axis.

$\therefore\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{a}$

$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{a}$

$\Rightarrow2\text{b}^2=\text{a}$

Eccentricity, $\text{e}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{\text{a}}$

Substituting the value $\text{b}^2=\frac{\text{a}^2}{2},$ we get:

$\text{e}=\frac{\sqrt{\text{a}^+\frac{\text{a}}{2}}}{\text{a}}$

$=\frac{\text{a}\sqrt{\frac{3}{2}}}{\text{a}}$

$=\sqrt{\frac{3}{2}}$

$\therefore$ Eccentricity is $\sqrt{\frac{3}{2}}$

3. $\sqrt{{\text{a}^2 × (1 + \text{m}^2)} > \text{c}}$

4. $\frac{2\sqrt{2}}{\sqrt{2}}$

2. $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$

Solution:

The equation of a hyperbola with foci on the x-axis is defined as. $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$

3. x² + y² - 4x - 2y + 4 = 0

Solution:

distance of pt. (2, 1) from line 3x + 4y - 5 is radius(r)

$\Rightarrow\text{r}=\frac{\mid6+4-5\mid}{5}=\frac{5}{5}=1$

⇒ Equation of circle is

⇒ (x - 2)² + (y - 1)² = 1

⇒ x² + y² - 4x - 2y + 4 = 0

1. $(-1,\ 3)$

Solution:

The given equation of the curve is x² + y² = 25

Since $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve x² + y² = 25 and the y-axis, we have:

$\lambda^2+(\lambda+1)^2<25,$ provided $\lambda+1>0$

$\Rightarrow\lambda^2+\lambda^2+12\lambda<25,\ \lambda>-1$

$\Rightarrow2\lambda^2+2\lambda-24<0,\ \lambda>-1$

$\Rightarrow\lambda^2+\lambda-12<0,\ \lambda>-1$

$\Rightarrow(\lambda-3)(\lambda+4)<0,\ \lambda>-1$

$\Rightarrow-4<\lambda<3,\ \lambda>-1$

$\Rightarrow\lambda\in(-1,\ 3)$