The perpendicular distance from the origin to the focal chord drawn through the point $(4,5)$ to the…

The perpendicular distance from the origin to the focal chord drawn through the point $(4,5)$ to the parabola $y^2-4 y-3 x+7=0$ is
  1. $\frac{2}{5}$
  2. $\frac{1}{\sqrt{2}}$
  3. $\frac{1}{5}$
  4. $1$

Solution

Given equation of parabola $\begin{aligned} & y^2-4 y-3 x+7=0 \Rightarrow(y-2)^2=3(x-1) \\ & \Rightarrow(y-2)^2=4 \cdot \frac{3}{4}(x-1) \end{aligned}$ So focus is $\left(\frac{3}{4}+1,2\right)=\left(\frac{7}{4}, 2\right)$ Now, equation of focal chord is $y-5=\frac{2-5}{\frac{7}{4}-4}(x-4) \Rightarrow 4 x-3 y-1=$ Distance between focal chord \& origin is $\frac{|0-0-1|}{\sqrt{(-3)^2-4^2}}=\frac{1}{5}$

Asked in: AP EAMCET 2023 (15 May Shift 2)

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