Straight Lines

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$

slope of HC = $\frac{16p-p^2-31}{16p-32}$

slope of BC * slope of HC = -1 p = 3 or 5

hence p = 3 is only possible value.

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P (h, k) is $x^2+y^2+x-3y-2=0$

Now intersection with x – axis are $x^2+x-2=0\Rightarrow x=-2,1$

Now intersection with y – axis are $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$

Therefore are of the quadrilateral ABCD is = $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f   $\left(\frac{1}{4}\right)f\left(\frac{1}{3}\right)<0\Rightarrow t\in \left(\frac{1}{4},\frac{1}{3}\right)$

Let P(1 – sin q, -1 + cos q) $\therefore$  slope of normal = slope of CP $-\frac{1}{2t}=\frac{cos\theta }{-sin\theta }\Rightarrow 2t$ Þ = tan q according to question $x=1-sin\theta =1-\frac{2t}{\sqrt[]{1+4t^2}}=g\left(t\right)\therefore g\left(t\right)=1-\frac{2t}{\sqrt[]{1+4t^2}}$ ,  

Þ g'(t) < 0 g(t) is decreasing function in  $t\in \left(\frac{1}{4},\frac{1}{3}\right)\Rightarrow g\left(t\right)\in \left(0.440,0.485\right)\in \left(\frac{1}{4},\frac{1}{2}\right)$

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\theta =\frac{10}{\mathrm{x}}=\frac{\mathrm{h}}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=\frac{\mathrm{h}\mathrm{x}}{10}$

$\mathrm{t}\mathrm{a}\mathrm{n}??=\frac{15}{\mathrm{x}}=\frac{\mathrm{h}}{\mathrm{x}}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{h}\mathrm{x}}{15}$

Now,  $x_1+x_2=x=\frac{hx}{15}+\frac{hx}{10}$

$\Rightarrow 1=\frac{\mathrm{h}}{10}+\frac{\mathrm{h}}{15}\Rightarrow \mathrm{h}=6$

1 < a₁ < a₂ ……18 < 77

77 = 1 + (20 – 1) . d If n numbers are in A.P.

76 = 19 * d ⇒ d = 4

⇒ a₁ = 5

$\begin{array}{l}\therefore a_1+a_2+....+a_{18}\\ =\frac{18}{2}\left[2*5+17*4\right]=702\end{array}$          

${\left(x-1\right)}^2+{\left(y-3\right)}^2=10-\alpha$

$\begin{array}{l}x-y=-1\\ x+y=a+b\end{array}\}\to m\left(\frac{a+b-1}{2},\frac{a+b+1}{2}\right)$

A' = 2m – A = (b – 1, a + 1)

$C_2:x^2+y^2+2gx+2fy+\frac{38}{5}=0$

r  $\sqrt[]{g^2+f^2-c}$

$=\sqrt[]{4+4-\frac{38}{5}}=\sqrt[]{\frac{2}{5}}$

$C_1:\sqrt[]{\frac{2}{5}}=\sqrt[]{1+9-\alpha }=\sqrt[]{10-\alpha }$

$\alpha +6r^2=\frac{48}{5}+6\left(\frac{2}{5}\right)=\frac{60}{5}=12$

 ${\left(a-1\right)}^2+4={\left(b-1\right)}^2+16$

$={\left(a-1\right)}^2+{\left(b-1\right)}^2$

${\left(b-1\right)}^2=4\&{\left(a-1\right)}^2=16$

$\Rightarrow b=1\pm 2a=1\pm 4$

= 3, 1= 5, 3

x + 2y = 3 …… (i)

3x – y = 8…… (ii)

x + 2y = 3

3x – y = 83 * 2

$\oplus \to 7x=-13\Rightarrow x=\frac{-13}{7}$

$y=\frac{-39}{7}+8=\frac{17}{7}$

$k_1+k_2=x+y=\frac{4}{7}$

y – 1 = m(x – 1)

mx – y + 1 – m = 0

$\frac{\left|1-m\right|}{\sqrt[]{m^2+1}}=\frac{4}{\sqrt[]{17}}$

$17{\left(1-m\right)}^2=16\left(m^2+1\right)$

m² – 3m + 1 = 0

$\begin{array}{l}bx+10y-8=0\\ y=\frac{2}{3}x\end{array}\}\Rightarrow \left(b+\frac{20}{3}\right)x=8$

$x=\frac{24}{3b+20},y=\frac{16}{3b+20}$

b = $\frac{2\left(3-2\sqrt[]{2}\right)\left(4+3\sqrt[]{2}\right)}{-2}$

$=\frac{12-8\sqrt[]{2}+9\sqrt[]{2}}{-1}=-\sqrt[]{2}$

b² = 2

$\frac{x^2}{5}+\frac{y^2}{2}=1$

2 = 5 (1 – e²)

$e^2=\frac{3}{5}$

$e$$=$

$D=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0\Rightarrow a=8$

also,  $D_1=\left|\begin{array}{ccc}9& -2& 3\\ b& 1& 1\\ 24& -7& 8\end{array}\right|=0\Rightarrow b=3$

hence,  $a-b=8-3=5$