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| 1. |
Match the following lists: |
Answer» (a) OBVIOUSLY, all the points in List II are common to the hyperbola and circle. (b). The chord of contact of hyperbola w.r.t. (0, -9/4) is `0(x)-(-(9)/(4))y=9` `"or"y=4` Solving this with hyperbola, we have `x^(2)-16=9` `"or"x^(2)=25or x= pm5` Hence, the point of contact are `(pm 5, 4)`. (c) Obviously, the required point is `(-5, -4)`. (d) LET the point on the hyperbola be P(H,k) and `Q(-h,k)`. Then `"Area of TRIANGLE"=(1)/(2)|2h||-6-k|=10"(1)"` Also, points P and Q lie on the hyperbola. Hence, `h^(2)-k^(2)=9"(2)"` Clearly, point `(pm5,-4)` satisfy both (1) and (2). |
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