The foci of the hyperbola are S(5,6),S′(−3,−2). If its eccentricity is 2, then the equation of its directrix corresponding to focus S is
A
x+y−3=0
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B
x+y−5=0
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C
x+y−7=0
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D
x+y−1=0
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Solution
The correct option is Bx+y−5=0
SS′=√(5+3)2+(6+2)2=8√2 ⇒SS′=2ae=8√2 ⇒a=2√2(∵e=2) ⇒ae=√2=OF SF=4√2−√2=3√2
Since, slope of axis SS′=−2−6−3−5=1 ∴ slope of directrix =−1
Equation of directrix is given by y=−x+c ⇒y+x−c=0
Now, 3√2=∣∣∣5+6−c√2∣∣∣(perpendicular distance of a point from a line) ⇒|11−c|=6 ⇒11−c=±6 ⇒c=5 or c=17
Hence, Equation of Directrix can be x+y−5=0 or x+y−17=0
centre(O)≡(1,2)
we know that centre and focus lies on opposite side to the directrix. But for x+y−17=0, centre(O) and focus(S) lies on the same side.Therefore x+y−17=0 is rejected. ∴ Equation of directrix corresponding to focus S is x+y−5=0