$$ \begin{aligned} & \mathrm{AB}:(\lambda+1) x+\lambda y=4 \\\\ & \mathrm{AC}: \lambda x+(1-\lambda) y+\lambda=0 \\\\ & \text { Vertex } A \text { is on } y \text {-axis } \\\\ & \Rightarrow x=0 \end{aligned} $$

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 52 English Explanation 1

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 52 English Explanation 1

So $\mathrm{y}=\frac{4}{\lambda}, \mathrm{y}=\frac{\lambda}{\lambda-1}$

$$ \begin{aligned} & \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} \\\\ & \Rightarrow \lambda=2 \\\\ & \mathrm{AB}: 3 x+2 y=4 \\\\ & \mathrm{AC}: 2 \mathrm{x}-\mathrm{y}+2=0 \\\\ & \Rightarrow \mathrm{A}(0,2) \text { Let } \mathrm{C}(\alpha, 2 \alpha+2) \end{aligned} $$

Now (Slope of Altitude through C) $\left(-\frac{3}{2}\right)=-1$

$$ \left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2} $$

So $\mathrm{C}\left(-\frac{1}{2}, 1\right)$

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 52 English Explanation 2

JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 52 English Explanation 2

Let Equation of tangent be $y=m x+\frac{3}{2 m}$

$$ \begin{aligned} & \mathrm{m}^2+2 \mathrm{~m}-3=0 \\\\ & \Rightarrow \mathrm{m}=1,-3 \end{aligned} $$

So tangent which touches in first quadrant at $\mathrm{T}$ is

$$ \begin{aligned} & \mathrm{T} \equiv\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right) \\\\ & \equiv\left(\frac{3}{2}, 3\right) \\\\ & \Rightarrow \mathrm{CT}=\sqrt{4+4}=2 \sqrt{2} \end{aligned} $$

Parabola

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