Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

2x + y – 3z = 4

${\pi }_2=\left|\begin{array}{ccc}\hfill x-2\hfill & \hfill y+3\hfill & \hfill z-2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right|=0$

${\pi }_2:-5x+5y+z+23=0$

now line lying in both the planes have DR.

$\frac{a}{1+15}=\frac{b}{15-2}=\frac{c}{10+5}$

$\frac{a}{16}=\frac{b}{13}=\frac{c}{15}$

So direction ratio's a : b : c = 16 : 13 : 15

$x+f\text{'}\left(x\right)=f\text{'}\left(0\right)$

$f\text{'}\left(0\right)=1-c^{2}fromequation\left(i\right)$

$x+f\text{'}\left(x\right)=1-c^2$

$\frac{x^2}{2}+f\left(x\right)=\left(1-c^2\right)x+d$

f(0) = 0

$f\left(x\right)=\frac{-x^2}{\alpha }+\left(1-c^2\right)x$

c = 3/2

$f"\left(x\right)=-1=2\left(c+1\right)$

$f\left(x\right)=\frac{-x^2}{2}+\left(\frac{-5}{4}\right)x$

now $\left(f\left(1\right)+f\left(2\right)+.......f\left(20\right)\right)$

$=\frac{-1}{2}\left(1^2+2^2+......+20^2\right)-\frac{5}{4}\left(1+2+....+20\right)$

$=\frac{20\cdot 21}{2}\cdot \frac{\left(-82-15\right)}{12}$

now $2\left|f\left(1\right)+f\left(2\right)+....+f\left(20\right)\right|$

$=\frac{20\cdot 21\cdot 97}{12}=3395$

Given 2a + 2b = $4\left(2\sqrt[]{2}+\sqrt[]{14}\right)$

$=4\sqrt[]{2}\left(2+\sqrt[]{7}\right)$

$b^2=a^2\left(\frac{11}{4}-1\right)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2}a$

Let $\pi \equiv -x+y+z-1=0$

1 2 > 0 as both pt.lies on same side

now $P_1=\left|\frac{-1+2-1-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

P2 = $\left|\frac{-2+\left(-1\right)+3-1}{\sqrt[]{3}}\right|=\frac{1}{\sqrt[]{3}}$

as P₁ = P₂ so distance between foot of perpendicular will be same as distance between the points

d = $\sqrt[]{{\left(1-2\right)}^2+{\left(2+1\right)}^2+{\left(-1-3\right)}^2}$

= $\sqrt[]{1+9+16}$$=$

Tangent to C₁ at (-1, 1) is T = 0                                                           

 x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping  from (3, 2) to centre

OP = $\left|\frac{3-2+2}{\sqrt[]{2}}\right|=\frac{3}{\sqrt[]{2}}$

AP = $\sqrt[]{r^2-O{P}^2}$

$=\sqrt[]{5-\frac{9}{2}}=\frac{1}{\sqrt[]{2}}$

$tan\theta =\frac{OP}{AP}=\frac{3/\sqrt[]{2}}{1/\sqrt[]{2}}=3$

area of $\Delta ABN=\frac{1}{2}A{N}^2\cdot sin2\theta$

AN = $\frac{\sqrt[]{5}}{3}$

$=\frac{1}{2}\cdot \frac{5}{9}\cdot \left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

$=\frac{5}{2.9}*\frac{2.3}{1+3^2}=\frac{1}{6}$

sin = $\frac{AP}{AN}$

 $AP\perp BP$

M₁ M₂ = 1

$\frac{2t}{t^2-3}*\frac{2/6}{3-\frac{1}{t^2}}=-1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

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

$4=\frac{2b^2}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b² = a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e}\left(1-e^2\right)$

e² + 2e – 1 = 0

$e=\frac{-2\pm \sqrt[]{4+4}}{2}$

$e=-1+\sqrt[]{2}$

now $\frac{1}{e^2}$$=$$3$$+$$2$

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2\left(6-4+28-2\right)}{3^2+{\left(-1\right)}^2+4^2}$

Let $\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-28}{13}=\lambda$

$x=3\lambda +2y=-\lambda +4z=4\lambda +7$

Now according to the question

(a, b, c) = $\left(3\lambda +2,-\lambda +4,4\lambda +7\right)$

Now $2a+b+2c=6\lambda +4-\lambda +4+8\lambda +14$

= $13\lambda +22$

= 28 + 22 [Use $\lambda =\frac{-28}{13}$ ]

= 6

$L_1:\mathcal{l}x-y+3\left(1-z\right)=1,x+2y-z=2$

plane containing the line P : 3x – 8y + 7z = 4

If $\overrightarrow{n}$ be vector parallel to L.

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \mathcal{l}\hfill & \hfill -1\hfill & \hfill 3\left(1-\mathcal{l}\right)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(6\mathcal{l}-5\right)\widehat{i}+\left(3-2\mathcal{l}\right)\widehat{j}+\left(2\mathcal{l}+1\right)\widehat{k}$ as P containing the line

$\therefore 3\left(6\mathcal{l}-5\right)-8\left(3-2\mathcal{l}\right)+7\left(2\mathcal{l}+1\right)=0$

$\Rightarrow \mathcal{l}=\frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5/3}{\sqrt[]{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt[]{83}}$

$\therefore 415co{s}^2\theta =125$

The normal vector to the plane is $\overline{n_1}*\overline{n_2}=\left|\begin{array}{ccc}\hfill i\hfill & \hfill 3\hfill & \hfill k\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill -a\hfill \end{array}\right|=\left(1-a\right)\widehat{i}+\widehat{j}+\widehat{k}$

$\therefore equationofplaneis\left(1-a\right)\left(x-1\right)+\left(y-1\right)+z=0$

(1 – a)x + y + z = 2 – a …… (i)

Now distance from (2, 1, 4) = $\sqrt[]{3}$

$\Rightarrow \sqrt[]{3}=\left|\frac{2\left(1-a\right)+1+4-\left(2-a\right)}{\sqrt[]{{\left(1-a\right)}^2+1+1}}\right|$

$\Rightarrow a^2+2a-8=0\Rightarrow a=-4,2$ the largest value of a = 2.

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$\therefore$ Equation of circle is $x^2+y^2-4x+2y-13=0$   ……………. (i)

But, given  $x^2+y^2+2ax+2by+c=0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12