Question

Two curves $C_1:y=x^2-3$ and $C_2:y=k{x}^2,k\in R$ intersect each other at two different points. The tangent drawn to $C_2$ at one of the points of intersection $A\equiv \left(a,y_1\right),(a>0)$ meets $C_1$ again at $B\left(1,y_2\right)$ $\left(y_1\ne y_2\right)$. The value of ' $a$ ' is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

$\text{ Point }A\left(a,y_1\right)\text{ lies on }{C}_1\text{ and }{C}_2$
$\text{ hence }{y}_1=a^2-3\text{ and }{y}_2=k{a}^2$
$\Rightarrow a^2-3=k{a}^2$
$\text{ now }y=k{x}^2\Rightarrow \frac{dy}{dx}=2kx$
${\therefore \frac{dy}{dx}]}_{\left(a,y_1\right)}=2ka=\frac{y_2-y_1}{1-a}\left(\text{ But }{y}_2=1-3=-2\right)$
$=\frac{-2-\left(a^2-3\right)}{1-a}\Rightarrow 2ka=\frac{1-a^2}{1-a}=1+a$
$2ka=1+a$
Substituting $k=\frac{a^2-3}{a^2}$ from (1) in (2) we get $\frac{2a\left(a^2-3\right)}{a^2}=1+a\Rightarrow 2a^2-6=a+a^2$ $\Rightarrow a^2-a-6=0\Rightarrow a=+3,a=-2$ (rejected) ]