Points from which two distinct tangents can be drawn to two different branches of the hyperbola $\displaystyle \frac{x^2}{25}-\frac{y^2}{16}=1$ but no two different tangent can be drawn to the circle
$x^2+y^2=36$ is $(a)\; (1,6)\;\;\;\; (b)\; (1,2)\;\; (c)\;\; (7,1)\;\;\;\; (d)(1,0.5)$
Try: for two distinct tangents on the hyperbola $\displaystyle \frac{x^2}{25}-\frac{y^2}{16}=1$ , points must lies in between two asymptotes of hyperbola , which is $\displaystyle y = \pm \frac{4}{5}x$, could some help me to solve it, thanks
Let us first consider the condition about the circle.
If $P(X,Y)$ is inside the circle, then there are no tangent lines from $P$ to the circle.
If $P(X,Y)$ is on the circle, then there is only one tangent line at $P$.
If $P(X,Y)$ is outside the circle, then there are two tangent lines from $P$ to the circle.
So, our condition about the circle is equivalent to that $P(X,Y)$ is either on or inside the circle, i.e. $$X^2+Y^2\le 36\tag1$$
Next, let us consider the condition about the hyperbola.
Let $(p,q)$ be a point on the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ where $$\frac{p^2}{25}-\frac{q^2}{16}=1\iff 16p^2-25q^2=16\times 25\tag2$$
Since $y'=\frac{16x}{25y}$, the equation of the tangent line at $(p,q)$ is given by $$y-q=\frac{16p}{25q}(x-p)$$ If $P(X,Y)$ is on the tanget line, we have $$Y-q=\frac{16p}{25q}(X-p),$$ i.e. $$25qY=16pX-(16p^2-25q^2),$$ Using here $(1)$ gives $$25qY=16pX-16\times 25=16(pX-25)\tag3$$
Squaring the both sides of $(3)$, we get $$25Y^2\times 25q^2=16^2(pX-25)^2$$ Using $(2)$, we have $$25Y^2\times (16p^2-16\times 25)=16^2(pX-25)^2,$$ i.e. $$(25Y^2-16X^2)p^2+(2\times 16\times 25\times X)p-25^2Y^2-16\times 25^2=0\tag4$$
Seeing this as a quadratic equation on $p$, our condition is equivalent to that $(4)$ has two distinct real solutions and the product of two real solutions is negative, i.e. $$(2\times 16\times 25\times X)^2-4(25Y^2-16X^2)(-25^2Y^2-16\times 25^2)\gt 0\tag5$$and $$\frac{-25^2Y^2-16\times 25^2}{25Y^2-16X^2}\lt 0\tag6$$
$(5)$ is equivalent to $$16^2X^2+(25Y^2-16X^2)(Y^2+16)\gt 0\tag7$$
$(6)$ is equivalent to $$25Y^2-16X^2\gt 0\tag8$$
Here, we can see that if $(8)$ holds, then $(7)$ holds.
So, our condition about the hyperbola is $$(8)\iff (5Y-4X)(5Y+4X)\gt 0\tag9$$
Therefore, all we need is to find the options satisfying both $(1)$ and $(9)$.
$(a)(1,6)$ does not satisfy $(1)$.
$(b)(1,2)$ satisfies both $(1)$ and $(9)$.
$(c)(7,1)$ does not satisfy $(1)$.
$(d)(1,0.5)$ does not satisfy $(9)$.
It follows that $\color{red}{(b)}$ is the only correct option.