Two curves C_1 : y = x^2 - 3 and C_2 : y = kx | vedclass

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(B) Given curves are $C_1: y = x^2 - 3$ and $C_2: y = kx^2$.Since point $A(a, y_1)$ lies on both curves,we have $y_1 = a^2 - 3$ and $y_1 = ka^2$.Equat

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$3$

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(B) Given curves are $C_1: y = x^2 - 3$ and $C_2: y = kx^2$.Since point $A(a, y_1)$ lies on both curves,we have $y_1 = a^2 - 3$ and $y_1 = ka^2$.Equating these,$ka^2 = a^2 - 3$,so $k = \frac{a^2 - 3}{a^2} = 1 - \frac{3}{a^2}$.Point $B(1, y_2)$ lies on $C_1$,so $y_2 = 1^2 - 3 = -2$.Thus,$B$ is $(1, -2)$.The tangent to $C_2$ at $A(a, y_1)$ is given by $\frac{dy}{dx} = 2kx$. At $x = a$,the slope is $m = 2ka$.The equation of the tangent at $A(a, y_1)$ is $y - y_1 = 2ka(x - a)$.Since this tangent passes through $B(1, -2)$,we have $-2 - y_1 = 2ka(1 - a)$.Substitute $y_1 = a^2 - 3$ and $k = \frac{a^2 - 3}{a^2}$:$-2 - (a^2 - 3) = 2 \left( \frac{a^2 - 3}{a^2} ight) a (1 - a)$$1 - a^2 = 2 \left( \frac{a^2 - 3}{a} ight) (1 - a)$$(1 - a)(1 + a) = \frac{2(a^2 - 3)(1 - a)}{a}$.Since $a > 0$ and $A$ is an intersection point,$a 
eq 1$ (otherwise $C_1$ and $C_2$ would be tangent at $x=1$,but $y_1 
eq y_2$ implies $B$ is a distinct point). Dividing by $(1 - a)$:$1 + a = \frac{2(a^2 - 3)}{a}$$a + a^2 = 2a^2 - 6$$a^2 - a - 6 = 0$$(a - 3)(a + 2) = 0$.Since $a > 0$,we get $a = 3$.

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

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