Two curves $C1 : y = x^2– 3$and  $C2 : y = k{x}^2,$$\mathrm{k}\in \mathrm{R}$ intersect each other at two different points. The tangent drawn to C₂ at one of the point of intersection $A = (a, y_1) (a > 0)$  meets C₁ again at $B(1, y_2)$$({\mathrm{y}}_1\ne {\mathrm{y}}_2)$. The value of ‘$a$’ is

Point A(a, y₁) lies on C₁ and C₂

            Hence y₁ = a² – 3 and y₂ = ka²

            $\Rightarrow$a² – 3 = ka² ………..(1)

            Now y = kx² $\Rightarrow \frac{\text{dy}}{\text{dx}}\text{=2kx}$$\therefore {\frac{\text{dy}}{\text{dx}}]}_{\left({\text{a,\hspace{0.17em}\hspace{0.17em}y}}_{\text{1}}\right)}$

            $\text{=2ka\hspace{0.17em}=}\frac{{\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}}{\text{1-a}}$ (But y₂ = 1 – 3 = -2)

            $\text{=\hspace{0.17em}}\frac{\text{-2-}\left({\text{a}}^{\text{2}}\text{-3}\right)}{\text{1-a}}$

            $\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}2ka=}\frac{{\text{1-a}}^{\text{2}}}{\text{1-a}}\text{\hspace{0.17em}=1+a}$

            2ka = 1 + a ………..(2)

Substituting $\text{k\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{a}}^{\text{2}}\text{-3}}{{\text{a}}^{\text{2}}}$

            From (1) in (2) we get

            $\frac{\text{2a}\left({\text{a}}^{\text{2}}\text{-3}\right)}{\text{2}}\text{=\hspace{0.17em}1+\hspace{0.17em}\hspace{0.17em}a}$

            $\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2a}}^{\text{2}}{\text{-6=a+a}}^{\text{2}}$

            $\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}^{\text{2}}\text{-a-6=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}=\hspace{0.17em}+3\hspace{0.17em},\hspace{0.17em}\hspace{0.17em}a=\hspace{0.17em}-2\hspace{0.17em}}\left(\text{rejected}\right)$