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  • Equation of parabola having the extremities of it's latus rectum as and <img alt="(4,3)" src="https://entra

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

Option: 1

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


Option: 4

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 'a' \mathrm{= \frac{1}{4}(\sqrt{1+1})=\frac{\sqrt{2}}{4}}. Let the equation of it's directrix be \mathrm{y+x+\lambda= 0}.

\mathrm{\Rightarrow \quad \frac{|3+4+\lambda|}{\sqrt{2}}=2 \cdot \frac{\sqrt{2}}{4} }
\mathrm{\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} }

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

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mansi

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