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  • Consider the parabola with vertex <img alt="\mathrm{\left(\frac{1}{2}, \frac{3}{4}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%2C%20%5C

Consider the parabola with vertex \mathrm{\left(\frac{1}{2}, \frac{3}{4}\right)} and the directrix \mathrm{y=\frac{1}{2}}. Let P be the point where the parabola meets the line \mathrm{x=-\frac{1}{2}}. If the normal to the parabola at P intersects the parabola again at the point Q, then \mathrm{(P Q)^2} is equal to

Option: 1

\frac{25}{2}


Option: 2

\frac{75}{8}


Option: 3

\frac{125}{16}


Option: 4

\frac{15}{2}


Answers (1)

best_answer

Length of perpendicular from vertex to directrix \mathrm{=a=\frac{1}{4}}

\mathrm{\text { Equation of parabola, is } y-\frac{3}{4}=\left(x-\frac{1}{2}\right)^2}            ...[i]

\mathrm{\Rightarrow \frac{d y}{d x}=\left.2\left(x-\frac{1}{2}\right) \Rightarrow \frac{d y}{d x}\right|_{\text {At } x=\frac{-1}{2}}=-2}

Slope of normal

\mathrm{\left(\text { at } x=\frac{-1}{2}\right)=\frac{1}{2}}

\mathrm{\text { When } x=\frac{-1}{2} \text {, }}

\mathrm{y-\frac{3}{4}=\left(\frac{-1}{2}-\frac{1}{2}\right)^2 \Rightarrow y=\frac{7}{4}}

\mathrm{\text { Equation of normal at }\left(\frac{-1}{2}, \frac{7}{4}\right) \text { with slope } \frac{1}{2} \text { is, }}

\mathrm{\begin{aligned} & y-\frac{7}{4}=\frac{1}{2}\left(x+\frac{1}{2}\right) \\ \Rightarrow & 4 y-7=2 x+1 \\ \Rightarrow & y=\left(\frac{x}{2}+2\right) \end{aligned}}               ...[ii]

\mathrm{\therefore \quad \text { For point of intersection, }}          (Using (i))

\mathrm{\begin{aligned} & \frac{x}{2}+2-\frac{3}{4}=x^2-x+\frac{1}{4} \\ \Rightarrow & x^2-\frac{3 x}{2}-1=0 \end{aligned}}

\mathrm{\text { Let } x_1, x_2 \text { be roots of it, then } x_1+x_2=\frac{3}{2} \text { and } x_1 x_2=-1}

\mathrm{\Rightarrow\left(x_1-x_2\right)^2=\frac{9}{4}+4=\frac{25}{4}}

By using (ii), we have 

\mathrm{\begin{aligned} & y_1=\frac{x_1}{2}+2, y_2=\frac{x_2}{2}+2 \\ & \therefore \quad\left(y_1-y_2\right)^2=\frac{\left(x_1-x_2\right)^2}{4}=\frac{25}{16} \end{aligned}}

\mathrm{\text { Thus, }\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2=P Q^2=\frac{25}{4}+\frac{25}{16}=\frac{125}{16}}

 

Posted by

HARSH KANKARIA

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