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Let \mathrm{L_1: x+y=0 and L_2: x-y=0} are tangent to a parabola whose focus is \mathrm{S(1,2)}

If the length of latus rectum of the parabola can be expressed at \mathrm{\frac{m}{\sqrt{n}}} (where \mathrm{m \: and \: n}are coprime), then find the value of \mathrm{\frac{4(m+n)}{11}=}

Option: 1

4


Option: 2

1


Option: 3

6


Option: 4

5


Answers (1)

best_answer

Feet of the perpendicular \mathrm{ \: \left(N_1\right.\, and \: \left.N_2\right)} from focus upon any tangent to parabola lies on the tangent line at the vertex.

Now equation of \mathrm{S N_1\: is \: x+y=\lambda} passing through \mathrm{(1,2)}
\mathrm{ \Rightarrow \lambda=3 }
Equation of \mathrm{ S N_1\: is \: x+y=3. }
Solving \mathrm{ x+y=3 and y=x, we \: get N_1 \equiv\left(\frac{3}{2}, \frac{3}{2}\right) }

Similarly equation of \mathrm{ \mathrm{SN}_2 \: is \: x-y=\lambda } passing through \mathrm{ (1,2) }

\mathrm{ \Rightarrow \lambda=-1 }

Equation of \mathrm{ \mathrm{SN}_2 \: is \: y-x=1 }

Solving \mathrm{ y-x=1\: and \: y=-x, }
we get \mathrm{ N_2 \equiv\left(-\frac{1}{2}, \frac{1}{2}\right) }

Now equation of tangents line at vertex is \mathrm{ 2 x-4 y+3=0. }

Distance of \mathrm{ S(1,2) } from tanat vertex is

\mathrm{ =\frac{|2-8+3|}{\sqrt{20}}=\frac{3}{2 \sqrt{5}}=\frac{1}{4} \times \text { latus rectum } }
and hence length of rectum \mathrm{ =\frac{6}{\sqrt{5}}=\frac{m}{\sqrt{n}} }

Hence, \mathrm{ m+n=6+5=11. }

Posted by

sudhir.kumar

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