A normal to the hyperbola x26−y22=1 has equal intercepts on positive x and y axes. If this normal touches the ellipse x2a2+y2b2=1, then the value of a2+b2 is
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answer is 16.
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Detailed Solution
The equation of normal to hyperbolax26−y22=1 at (6secθ,2tanθ) is 6xsecθ+2ytanθ=8 Slope =−1⇒ −6secθ×tanθ2=−1⇒ sinθ=13 Thus, equation of normal is x+y=4 So, y=−x+4 is tangent to x2a2+y2b2=1 Hence, a2+b2=16