Question

 If \[{{F}_{1}}\] and \[{{F}_{2}}\] are the foci of a hyperbola, and P is a point on one of its branches, then the tangent to the curve at P, bisects the angle ${{F}_{1}}P{{F}_{2}}$ in the ratio

a) 2:1

b) 1:1 

c) 3:1 

d) 4:1

Answer 1

Hint: Now first we will understand the condition required to form a hyperbola and try to use the equation to form the equation of hyperbola. Now we will use the property of hyperbola which gives us a relation between tangent and the angle formed by joining the point on hyperbola to the foci of hyperbola.

Complete step-by-step solution:
Now let us consider the figure of Hyperbola.

Now in the above figure ${{F}_{1}}$ and ${{F}_{2}}$ are the foci of the Hyperbola. O is the origin and the lines passing through A and A’ are the directrix of the hyperbola.

Now let the point P be $\left( x,y \right)$ .

Hence we can say that $OS=x$ and $PS=y................\left( 1 \right)$

Now we know that the eccentricity of the conic section is the ratio of the distance between any point on conic to the focus and perpendicular distance from that point to the directrix.

Hence we have $e=\dfrac{P{{F}_{1}}}{PP'}$

Now we will try to write the equation of hyperbola from the definition.

Now we know that PP’ is nothing but equal to AS’.

Hence we get $P{{F}_{1}}=eAS’..............\left( 2 \right)$

Now AS is nothing but $OS-OA'$

Now substituting this in equation (2) we get,

$\Rightarrow P{{F}_{1}}=e\left( OS-OA' \right)$

Now squaring the above equation on both sides we get,

$\Rightarrow P{{F}_{1}}^{2}={{e}^{2}}{{\left( OS-OA' \right)}^{2}}$

Now we know that the distance of the directrix from the origin is given by $\dfrac{a}{e}$ where $\left( a,0 \right)$ is the vertex of the hyperbola. Hence we get $OA'=\dfrac{a}{e}$ and we know that OS = x from (1).

Hence we get,

$\Rightarrow P{{F}_{1}}^{2}={{e}^{2}}{{\left( x-\dfrac{a}{e} \right)}^{2}}$

Now using Pythagoras theorem in triangle $P{{F}_{1}}S$ we get,

$\Rightarrow \left( P{{S}^{2}}+{{F}_{1}}{{S}^{2}} \right)={{e}^{2}}{{\left( x-\dfrac{a}{e} \right)}^{2}}$

Now from equation (1) we have PS = y and we know that ${{F}_{1}}S=OS-O{{F}_{1}}$. Now we have OS = x and $O{{F}_{1}}=ae$. Hence using this in the equation we get,

$ \Rightarrow {{\left( y \right)}^{2}}+{{\left( x-ae \right)}^{2}}={{e}^{2}}{{\left( x-\dfrac{a}{e} \right)}^{2}} $

$  \Rightarrow {{y}^{2}}+{{x}^{2}}+{{a}^{2}}{{e}^{2}}-2aex={{\left( xe-a \right)}^{2}} $

$ \Rightarrow {{x}^{2}}+{{y}^{2}}+{{a}^{2}}{{e}^{2}}-2aex={{x}^{2}}{{e}^{2}}+{{a}^{2}}-2eax $

$ \Rightarrow {{x}^{2}}-{{x}^{2}}{{e}^{2}}+{{y}^{2}}={{a}^{2}}-{{a}^{2}}{{e}^{2}} $

$  \Rightarrow {{x}^{2}}\left( 1-{{e}^{2}} \right)+{{y}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) $

Now dividing the whole equation by \[\left( 1-{{e}^{2}} \right)\] we get,

$\Rightarrow {{x}^{2}}+\dfrac{{{y}^{2}}}{1-{{e}^{2}}}={{a}^{2}}$

Now dividing the whole equation by ${{a}^{2}}$ we get,

$\Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{a}^{2}}\left( 1-{{e}^{2}} \right)}=1$

Now we know that ${{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right)$.

Hence the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Now we know that for hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ the tangent drawn at any point bisects the angle between the lines, joining the point to the foci. Hence we can say that both the angles are the same.

Hence we can say that the ratio is 1 : 1. Therefore, the correct answer is option (b).

Note: Now note that the tangent drawn at any point bisects the angle between the line made by joining the point to the two foci of hyperbola, whereas the normal bisects the supplementary angle between the line made by joining foci and point on Hyperbola.