The point from which two distinct tangents ca | vedclass

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Question

(B) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the region from which two tangents can be drawn to two different branches is the region

Vedclass · Accepted Answer

$(1, 3)$

Vedclass · Answer

(B) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the region from which two tangents can be drawn to two different branches is the region between the two asymptotes $y = \pm \frac{b}{a}x$. Here,$a^2 = 25$ and $b^2 = 16$,so the asymptotes are $y = \pm \frac{4}{5}x$. For a point $(x_1, y_1)$ to be in this region,it must satisfy $\left| \frac{y_1}{x_1} \right| > \frac{4}{5}$. Checking the options: $(1, 6) \implies |6/1| = 6 > 0.8$ (True) $(1, 3) \implies |3/1| = 3 > 0.8$ (True) $(7, 1) \implies |1/7| \approx 0.14 $(1, 0.5) \implies |0.5/1| = 0.5 Now,for the circle $x^2 + y^2 = 36$,a point $(x_1, y_1)$ lies outside the circle if $x_1^2 + y_1^2 > 36$. If a point is outside,two tangents can be drawn. For $(1, 6)$,$1^2 + 6^2 = 37 > 36$. Thus,two tangents can be drawn to the circle. For $(1, 3)$,$1^2 + 3^2 = 10 Therefore,the correct point is $(1, 3)$.

The point from which two distinct tangents can be drawn to two different branches of the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$,but no two different tangents can be drawn to the circle $x^2 + y^2 = 36$,is:

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