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COLUMN - I COLUMN - II
A) The points common to the hyperbola $x^{2} - y^{2} = 9$ and the circle $x^{2} + y^{2} = 41$ are p) (–5, –4)
B) Tangents drawn from point $(0, - \frac{9}{4})$ to the hyperbola $x^{2} - y^{2} = 9$ , then the point of tangency may have coordinate(s) q) (5, 4)
C) The point which is diametrically opposite of point (5, 4) with respect to the hyperbola $x^{2} - y^{2} = 9$ is r) (–5, 4)
D) If P and Q lie on the hyperbola $x^{2} - y^{2} = 9$ such that area of the isosceles triangle PQR where PR = QR is 10 sq. units, where R=(0, 6), then P can have the coordinate(s) s) (5, –4)

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A-pq; B-qr; C-p; D-pq

A-pqrs; B-qr; C-p; D-ps

A-rs; B-qr; C-p; D-q

A-p; B-pqr; C-p; D-ps

answer is C.

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Detailed Solution

$A) Solve x^{2} - y^{2} = 9, x^{2} + y^{2} = 41,$

$∴ P (x, y) = (\pm 5, \pm 4)$

B) Equation of chord of contact of $(0, \frac{- 9}{4})$ to $x^{2} - y^{2} = 9$ is $x (a) - y (\frac{- 9}{4}) = 9 \Rightarrow y = 4$

Solve $solve y = 4 and x^{2} - y^{2} = 9 w e g e t P (x, y) = (\pm 5, 4)$

$C) P (5, 4), C (0, 0)$ Opposite point = (-5, -4)

$D) Let P = (h, k), Q (- h, k)$

$Area = Δ = \frac{1}{2} | 2 h | | - 6 - k | = 10 \Rightarrow | h | | 6 + k | = 10$ ….(1)

$(h, k) lies on x^{2} - y^{2} = 9$

$\Rightarrow h^{2} - k^{2} = 9$ …..(2)

$\Rightarrow Solve (1) & (2) \Rightarrow (\pm 5, - 4)$

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