Question

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is –
(a) $2x-y-2=0$
(b) $3x-2y=0$
(c) $2x-3y+10=0$
(d) $2x-2y+8=0$

Answer 1

To find the equation of tangent of hyperbola at a particular point we must know the general equation of hyperbola. We will use the relation ${{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}\,$ to find the value of ${{a}^{2}}$ and ${{b}^{2}}$ . In this question to find these constants we have to put the given point in the general equation of hyperbola and use the eccentricity equation of hyperbola.

Complete step by step answer:
In the question it is given that the hyperbola passes through the point (4,6) and eccentricity is 2.
We know that, the equation of hyperbola is : $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\,...(i)$

As the hyperbola is passing through the point (4,6) ,So this point must satisfy the equation $(i)$ .
So, putting the point (4,6) in equation $(i)$ ,we get –

& \dfrac{{{4}^{2}}}{{{a}^{2}}}-\dfrac{{{6}^{2}}}{{{b}^{2}}}=1 \\\ & \dfrac{16}{{{a}^{2}}}-\dfrac{36}{{{b}^{2}}}=1\,...(ii) \\\ \end{aligned}$$ Now ,the eccentricity equation of the hyperbola is : $${{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}\,\,...(iii)$$ And it is given that the eccentricity of the given hyperbola is 2. So putting the value of eccentricity i.e. e = 2 in equation $$(iii)$$ , we get – $$\begin{aligned} & {{2}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\\ & 4=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\\ & \Rightarrow {{b}^{2}}=3{{a}^{2}}\,...(iv) \\\ \end{aligned}$$ Now, putting the value of $${{b}^{2}}$$ from equation $$(iv)$$ in equation $$(ii)$$ , we get – $$\begin{aligned} & \dfrac{16}{{{a}^{2}}}-\dfrac{36}{3{{a}^{2}}}=1 \\\ & \Rightarrow \dfrac{48-36}{3{{a}^{2}}}=1 \\\ & \Rightarrow {{a}^{2}}=4\,...(v) \\\ \end{aligned}$$ Now, putting the value of $${{a}^{2}}$$ from equation $$(v)$$ in equation $$(iv)$$ , we get – $$\begin{aligned} & {{b}^{2}}=3(4) \\\ & \Rightarrow {{b}^{2}}=12\,...(vi) \\\ \end{aligned}$$ Now, Putting the value of $${{a}^{2}}$$ and $${{b}^{2}}$$ from equation $$(v)$$ and equation $$(vi)$$ respectively in equation $$(i)$$, we get – $$\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{12}=1$$  This above equation is the equation of hyperbola. So, the equation of tangent of any hyperbola at point (x1, y1) is given by – $$\dfrac{{{x}_{1}}x}{{{a}^{2}}}-\dfrac{{{y}_{1}}y}{{{b}^{2}}}=1\,...(vii)$$ So to find the tangent equation put the point (x1, y1) =(4,6) in equation $$(vii)$$ ,we get – $$\begin{aligned} & \dfrac{4x}{4}-\dfrac{6y}{12}=1 \\\ & \Rightarrow x-\dfrac{y}{2}=1 \\\ & \Rightarrow 2x-y-2=0 \\\ \end{aligned}$$ **So, the correct option is option (a).** **Note:** The alternative method to do this is by eliminating the method as the tangent point is given it should satisfy the tangent equation. So, put the point (4,6) in the option and check whether it’s true or not. By doing this Option (d) is eliminated then convert all remaining options in the form of general equation of tangent which is – $$\dfrac{{{x}_{1}}x}{{{a}^{2}}}-\dfrac{{{y}_{1}}y}{{{b}^{2}}}=1$$ By doing this you will get the value of $${{a}^{2}}$$ and $${{b}^{2}}$$ , then put the value of $${{a}^{2}}$$ and $${{b}^{2}}$$ in general equation of hyperbola. Then this equation will satisfy the point which is lying on the hyperbola which is (4,6). The option which will satisfy is the correct option. In this case it is option (d).