The equations of the sides AB and AC of a tri | vedclass

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(B) The equations of the sides are $AB: (\lambda+1)x + \lambda y = 4$ and $AC: \lambda x + (1-\lambda)y + \lambda = 0$.Since vertex $A$ lies on the $y

Vedclass · Accepted Answer

$2 \sqrt{2}$

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(B) The equations of the sides are $AB: (\lambda+1)x + \lambda y = 4$ and $AC: \lambda x + (1-\lambda)y + \lambda = 0$.Since vertex $A$ lies on the $y$-axis,we set $x=0$ in both equations to find the $y$-coordinate of $A$.For $AB$,$y = 4/\lambda$. For $AC$,$y = \lambda/(\lambda-1)$.Equating these,$4/\lambda = \lambda/(\lambda-1)$ $\Rightarrow 4\lambda - 4 = \lambda^2$ $\Rightarrow \lambda^2 - 4\lambda + 4 = 0$ $\Rightarrow (\lambda-2)^2 = 0$ $\Rightarrow \lambda = 2$.Substituting $\lambda=2$,we get $AB: 3x + 2y = 4$ and $AC: 2x - y + 2 = 0$. Thus,$A$ is $(0,2)$.Let $C$ be $(\alpha, 2\alpha+2)$ (since $C$ lies on $AC$).The orthocentre $H(1,2)$ is the intersection of altitudes. The altitude from $C$ is perpendicular to $AB$. The slope of $AB$ is $-3/2$,so the slope of the altitude from $C$ is $2/3$.The line passing through $H(1,2)$ and $C(\alpha, 2\alpha+2)$ has slope $(2\alpha+2-2)/(\alpha-1) = 2\alpha/(\alpha-1)$.Setting $2\alpha/(\alpha-1) = 2/3$ $\Rightarrow 6\alpha = 2\alpha - 2$ $\Rightarrow 4\alpha = -2$ $\Rightarrow \alpha = -1/2$.Thus,$C$ is $(-1/2, 1)$.The parabola is $y^2 = 6x$,so $4a = 6 \Rightarrow a = 3/2$. The tangent is $y = mx + a/m = mx + 3/(2m)$.Since the tangent passes through $C(-1/2, 1)$,$1 = m(-1/2) + 3/(2m)$ $\Rightarrow 2 = -m + 3/m$ $\Rightarrow m^2 + 2m - 3 = 0$.Solving for $m$,$(m+3)(m-1) = 0 \Rightarrow m = 1$ or $m = -3$.For the first quadrant,the point of contact $T(a/m^2, 2a/m) = (3/(2m^2), 3/m)$.For $m=1$,$T = (3/2, 3)$. The distance $CT = \sqrt{(3/2 - (-1/2))^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.

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