Question

Let $m_{1,}{m}_2$ are slopes of two tangent that are drawn from $(2,3)$ to the parabola $y^2=4x$ and $\alpha$ is the harmonic mean of $m_1$&$m_2$ if $\beta$ is the value of $\left(1+\tan {23}^{\circ }\right)\left(1+\tan {22}^{\circ }\right)$ ,then $\frac{3}{2}\alpha +\beta$ is

Answer 1

$(1+\tan {23}^o)(1+\tan ({45}^o-{23}^o\left)\right)$

$=(1+\tan {23}^o)(1+\frac{\tan {45}^o-\tan {23}^o}{1+\tan {45}^o\tan {23}^o})$

$\left(\tan \right(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B})$

$=(1+\tan {23}^o)(1+\frac{1-\tan {23}^o}{1+\tan {23}^o})$ $(\tan {45}^o=1)$

$=(1+\tan {23}^o)\frac{(1+\tan {23}^o+1-\tan {23}^o)}{(1+\tan {23}^0)}$

$=2=\beta$

If $m_1$ & $m_2$ are slopes of tangents,

we know they intersect

at $\left(a\right(m_1+m_2),a{m}_1{m}_2)$ where

$y^2=4ax$ is eqn of parabola

Here, $a=1$

$\therefore m_1+m_2=2,m_1{m}_2=3$

Harmonic mean of $m_1$ & $m_2$

$=\frac{2}{\frac{1}{m_1}+\frac{1}{m_2}}=\frac{2}{\frac{m_2+m_1}{m_1{m}_2}}$

$=\frac{2m_1{m}_2}{m_1+m_2}=\frac{2\times 3}{2}$

$=3=\alpha$

$\therefore \frac{3}{2}\alpha +\beta =\frac{3\times 3}{2}+2=\frac{13}{2}$