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Equation of parabola having the extremities of it's latus rectum as (3,4) \text{and} (4,3)     is
 

Option: 1

\mathrm{\left(x-\frac{7}{3}\right)^2+\left(y-\frac{7}{3}\right)^2=\left(\frac{x+y-6}{2}\right)^2}


 


Option: 2

\mathrm{\left(x-\frac{7}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=\left(\frac{x+y-8}{2}\right)^2}
 


Option: 3

\mathrm{\left(x-\frac{7}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=\left(\frac{x+y-4}{2}\right)^2}
 


Option: 4

\text{None of these}


Answers (1)

best_answer

Focus is \left(\frac{7}{2}, \frac{7}{2}\right) and it's axis is the line \mathrm{y=x}. Corresponding value of \mathrm{' a '} is \mathrm{\frac{1}{4} \sqrt{(1+1)}=\frac{\sqrt{2}}{4}.}
Let the equation of it's directrix be \mathrm{\mathrm{y}+\mathrm{x}+\lambda=0 . \Rightarrow \frac{|3 \mathrm{x}+4+\lambda|}{\sqrt{2}}=2 . \frac{\sqrt{2}}{4} \Rightarrow \lambda=-6,-8}

Thus equation of parabola is

\mathrm{ \left(x-\frac{7}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=\frac{(x+y-6)^2}{2} }

\mathrm{\text { or } \quad \left(x-\frac{7}{2}\right)^2+\left(y-\frac{7}{2}\right)^2=\frac{(x+y-8)^2}{2} }

Hence option 2 is correct.


 

Posted by

Divya Prakash Singh

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