Question

Normals are drawn from a point P(h, k) with slopes $m_1,m_2,m_3$ to the parabola $C_1:y^2=4x$

${\sum }_{\alpha =0}^{\infty }(m_1{)}^{\alpha }+{\sum }_{\alpha =0}^{\infty }(m_2{)}^{\alpha }+{\sum }_{\alpha =0}^{\infty }(m_3{)}^{\alpha }=\frac{89}{22},where|m_i|<1\forall i=1,2,3then67h-89k=$

• A. 139
• B. 157
• C. 173
• D. 191

Answer 1

The correct option is B

157

$y=mx-2m-m^3{m}^3+m(2-h)+k=0$
${\sum }_{\alpha =0}^{\infty }(m_1{)}^{\alpha }+{\sum }_{\alpha =0}^{\infty }(m_2{)}^{\alpha }+{\sum }_{\alpha =0}^{\infty }(m_3{)}^{\alpha }=\frac{89}{22}$
$\frac{1}{1-m_1}+\frac{1}{1-m_2}+\frac{1}{1-m_3}=\frac{89}{22}{m}^3+(2-h)m+k=0\{\begin{array}{c}{m}_1\\ m_2\\ m_3\end{array}$
Put $m=\frac{t-1}{t}$
$\Rightarrow (t-1{)}^3+(2-h\left)\right(t-1)t^2+k{t}^3=0$
This equation has roots
$\frac{1}{1-m_1},\frac{1}{1-m_2},\frac{1}{1-m_3}$
Sum of roots $=\frac{89}{22}$
$\therefore \frac{5-h}{k-h+3}=\frac{89}{22}\Rightarrow 67h-89k=157$