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JEE MATHEMATICS MCQ: HYPERBOLA

1 Easy
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Application +4 | -1
Locus of the point of intersection of the tangents at the points with eccentric angles ϕ\phi and π2−ϕ\frac{\pi}{2}-\phi on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is
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Correct option: B
y=by=b
Option B (y=by=b) is correct because the locus of the point of intersection of tangents at the given eccentric angles simplifies to y=b after substituting the angles into the intersection point formula.

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