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  • Find the locus of a point from where a normal can be drawn to the parabola which makes an angle of 45o in anticlockwise direction with the +x-axis<strong

Find the locus of a point from where a normal can be drawn to the parabola which makes an angle of 45o in anticlockwise direction with the +x-axis

Option: 1

x-y-3a=0


Option: 2

x+y-3a=0


Option: 3

x-y+3a=0


Option: 4

x+y+3a=0


Answers (1)

best_answer

Let point be P(h,k) and let m1, m2, m3 be slopes of normals drawn from this point

Thus m1= tan 45o =1

Now put the value of m1=1 in relations derived for co-normal points

\begin{array}{l}{\mathrm{m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}=0} \Rightarrow \mathrm{m}_{2}+\mathrm{m}_{3} =-1 \\\\ {\mathrm{m}_{1} \mathrm{m}_{2}+\mathrm{m}_{2} \mathrm{m}_{3}+\mathrm{m}_{3} \mathrm{m}_{1}=\frac{2 \mathrm{a}-\mathrm{h}}{\mathrm{a}}} \Rightarrow { \mathrm{m}_{2}+\mathrm{m}_{2} \mathrm{m}_{3}+\mathrm{m}_{3}=\frac{2 \mathrm{a}-\mathrm{h}}{\mathrm{a}}} \\\\ {\mathrm{m}_{1} \mathrm{m}_{2} \mathrm{m}_{3}=-\frac{\mathrm{k}}{\mathrm{a}}\Rightarrow \mathrm{m}_{2} \mathrm{m}_{3}=-\frac{\mathrm{k}}{\mathrm{a}}}\end{array}

Solving the above three equation

\frac{ \mathrm{h}-2\mathrm{a}}{a}=1+\frac{k}{a}

h-k=3a

or x-y-3a=0

Posted by

SANGALDEEP SINGH

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