The locus of a point, from where pair of tangents to the rectangular hyperbola x2−y2=a2 contain an angle of 45∘, is :
A
(x2+y2)2+4a2(x2−y2)=4a4
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B
(x2+y2)2+4a2(x2−y2)=a4
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C
2(x2+y2)+4a2(x2−y2)=4a2
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D
(x2+y2)+a2(x2−y2)=4a2
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Solution
The correct option is A(x2+y2)2+4a2(x2−y2)=4a4 Equation of tangent to given hyperbola is : y=mx±√a2m2−a2 ⇒ Let P(h,k) be the locus ⇒k−mh=±√a2m2−a2 ⇒k2+m2h2−2kmh=a2m2−a2 ⇒m2(h2−a2)−2hkm+k2+a2=0 ∴m1+m2=2hkh2−a2 and m1m2=k2+a2h2−a2 Now, tan45∘=∣∣∣m1−m21+m1m2∣∣∣ ⇒(m1−m2)2=(1+m1m2)2 ⇒(m1+m2)2−4m1m2=(1+m1m2)2 ⇒(2hkh2−a2)2−4(k2+a2h2−a2)=(h2+k2h2−a2)2 ⇒(h2+k2)2+4a2(h2−k2)=4a4