Ncert Solutions Maths class 11th

Locus of point of intersection of perpendicular tangent will be its director circle & director circle of parabola be its directrix.

Given parabola y² = 16 (x – 3) so equation of directrix be x – 3 = - 4 i.e., x + 1 = 0

$y\left(x\right)=co{t}^{-1}\left(\frac{\sqrt[]{1+sinx}+\sqrt[]{1-sinx}}{\sqrt[]{1+sinx}-\sqrt[]{1-sinx}}\right),x\in \left(\frac{\pi }{2},\pi \right)$

$=co{t}^{-1}\left(\frac{sin\frac{x}{2}+cos\frac{x}{2}+sin\frac{x}{2}-cos\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}-sin\frac{x}{2}+cos\frac{x}{2}}\right)=co{t}^{-1}tan\frac{x}{2}=co{t}^{-1}cot\left(\frac{\pi }{2}-\frac{\pi }{2}\right)=\frac{\pi }{2}-\frac{x}{2}$

$\frac{dy}{dx}=\frac{-1}{2}$

${\left(3x^2+4x+3\right)}^2-\left(k+1\right)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)+k{\left(3x^2+4x+2\right)}^2=0$

 Let $3x^2+4x+3=a\&3x^2+4x+2=b\Rightarrow a-1$  

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D\ge 0$  

$16{\left(k-1\right)}^2-12\left(k-1\right)\left(2k-3\right)\ge 0$     

So, $k\in \left(1,\frac{5}{2}\right]$  

(y – 2)² = (x – 1)

2 (y – 2)    $\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}\left(2,3\right)=\frac{1}{2\left(3-2\right)}=\frac{1}{2}$              

Equation of tangent at P (2, 3):

$y-3=\frac{1}{2}\left(x-2\right)$  

2y – 6 = x – 2

x – 2y + 4 = 0

Q (-4, 0)

Required area = ${\displaystyle \underset{0}{\overset{3}{\int }}\left({\left(y-2\right)}^2+1-\left(2y-4\right)\right)dy=9}$

For divisibility by 5 last digit must be 0 or 5 but 0 is not possible in palindrome

so it will be 5.

So, required no. = 10 * 10 = 100

Equation of tangent to given ellipse at

$P:\frac{x.cos\theta }{b}+\frac{y.sin\theta }{2a}=1$

A (b sec θ. 0) 7 B (0, 2a cosec θ)

area of    $\Delta OAB$

$=\frac{2ab}{2sin\theta cos\theta }=\frac{2ab}{sin2\theta }$

For minimum area sin 2θ = 1

So minimum area = 2ab

=>k = 2

  $\underset{x\to \infty }{lim}\left(\sqrt[]{x^2-x+1}-ax\right)=b$

$\underset{x\to \infty }{lim}\left|\frac{x^2-x+1-a^2{x}^2}{\sqrt[]{x^2-x+1}+ax}\right|=b$

For existence of limit 1 – a² = 0 i.e. a = 1 only

$\underset{x\to \infty }{lim}\frac{1-x}{\sqrt[]{x^2-x+1}+x}=b$

$\Rightarrow b=\frac{-1}{2}$

So, (a, b) =   $\left(1,-\frac{1}{2}\right)$

$r=\sqrt[]{{\left(\frac{p}{2}\right)}^2+{\left(\frac{1-p}{2}\right)}^2-5}=\sqrt[]{\frac{p^2+1+p^2-2p-20}{4}}=\sqrt[]{\frac{2p^2-2p-19}{2}}$

Since $r\in \left(0,5\right)so0<\sqrt[]{2p^2-2p-19}<10$

$\Rightarrow 2p^2-2p-19>0\&2p^2-2p-19<100$     

$P\in \left[\frac{1-\sqrt[]{239}}{2},\frac{1-\sqrt[]{39}}{2}\right)\cup \left(\frac{1+\sqrt[]{39}}{2},\frac{1+\sqrt[]{239}}{2}\right]$

So number of integral values of P² is 61.

  $ta{n}^{-1}\frac{1}{2r^2}=ta{n}^{-1}\frac{2}{1+\left(4r^2-1\right)}$             

  $=ta{n}^{-1}\frac{\left(2r+1\right)-\left(2r-1\right)}{1+\left(2r+1\right)\left(2r-1\right)}$

${\displaystyle \sum _{r=1}^{50}\left[ta{n}^{-1}\left(2r+1\right)-ta{n}^{-1}\left(2r-1\right)\right]}$

$=ta{n}^{-1}101-ta{n}^{-1}1=ta{n}^{-1}\frac{100}{102}$

$\Rightarrow tanP=\frac{100}{102}=\frac{50}{51}$             

$P\left(-2\sqrt[]{6},\sqrt[]{3}\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$              

$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1.........\left(i\right)$              

$b^2=\frac{a^2}{4}.........\left(ii\right)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$  

Cuts conjugate axis at $R\left(0,R\sqrt[]{3}\right)$     

$\therefore QR=6\sqrt[]{3}$