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Q.Find the equation to the hyperbola whose foci are (4,2)(4, 2) and (8,2)(8, 2) and eccentricity is 2.

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 4mImportance★★★★★
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Centre (6,2)(6,2); 2ae=42ae=4 with e=2e=2 gives a=1a=1, b2=a2(e2−1)=3b^2=a^2(e^2-1)=3; transverse axis horizontal.

The foci (4,2)(4,2) and (8,2)(8,2) have the same yy-coordinate, so the transverse axis is horizontal and the centre is their midpoint:

Centre=(4+82,2+22)=(6,2).\text{Centre}=\left(\dfrac{4+8}{2},\dfrac{2+2}{2}\right)=(6,2).

Distance between the foci =2ae=∣8−4∣=4=2ae=|8-4|=4. With e=2e=2:

2a(2)=4⇒a=1.2a(2)=4\Rightarrow a=1.

b2=a2(e2−1)=1(4−1)=3.b^2=a^2(e^2-1)=1(4-1)=3.

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