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If the equation of the parabola, whose vertex is at \mathrm{\left ( 5,4 \right )} and the directrix is \mathrm{3 x+y-29=0, \text { is } x^{2}+a y^{2}+b x y+c x+d y+k=0} , then \mathrm{a+b+c+d+k} is equal to

Option: 1

575


Option: 2

-575


Option: 3

576


Option: 4

-576


Answers (1)

best_answer

Calculate foot of directrix

\mathrm{\frac{x-5}{3}=\frac{y-4}{1}=-1 \frac{(3 \times 5+4-29)}{3^{2}+1^{2}}} \\

\mathrm{\Rightarrow x=8, y=5 . }

Vertex is mid-point of focus and foot at directrix so coordinates of focus \mathrm{=\left ( 2,3 \right )}

Using \mathrm{PS=PM}, equation of parabola is

\mathrm{\sqrt{(x-2)^{2}+(y-3)^{2}}=\left|\frac{3 x+y-29}{\sqrt{10}}\right|}\\

\mathrm{\Rightarrow 10\left[x^{2}-4 x+4+y^{2}-6 y+9\right]=9 x^{2}+y^{2}+841+6 x y-58 y-174 x}\\

\mathrm{\Rightarrow x^{2}+9 y^{2}-6 x y+134 x-2 y-711=0 \text {. }}\\

Compare with  \mathrm{x^{2}+a y^{2}+b x y+c x+d y+x=0}\\

\mathrm{a+b+c+d+k: 9-6+134-2-711=143-719}\\

\mathrm{= { - 576 } }

Hence the correct answer is option 4.

Posted by

Divya Prakash Singh

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