Question

If ${{F}_{1}}\And {{F}_{2}}$ are the feet of the perpendiculars from foci ${{S}_{1}}\And {{S}_{2}}$ of an ellipse $\dfrac{{{x}^{2}}}{5}+\dfrac{{{y}^{2}}}{3}=1$ on the tangent at any point P on the ellipse then $\left( {{S}_{1}}{{F}_{1}} \right)\left( {{S}_{2}}{{F}_{2}} \right)=$ ?
(a) 8
(b) 5
(c) 3
(d) 9

Answer 1

We know that the coordinates of the foci of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to $\left( ae,0 \right)\And \left( -ae,0 \right)$ where “e” is the eccentricity and the formula for the eccentricity of this ellipse is equal to $1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$. Then we know that equation of a tangent at a point with slope “m is equal to $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$. In the above question, we asked to find the result of $\left( {{S}_{1}}{{F}_{1}} \right)\left( {{S}_{2}}{{F}_{2}} \right)$ which is equal to the product of the length of perpendicular drawn from the two foci ${{S}_{1}}\And {{S}_{2}}$ respectively.

Complete step by step solution:
In the above problem, we have given an ellipse which is equal to:
$\dfrac{{{x}^{2}}}{5}+\dfrac{{{y}^{2}}}{3}=1$
And also given that two perpendiculars are drawn from the two foci ${{S}_{1}}\And {{S}_{2}}$ and the feet of perpendiculars are ${{F}_{1}}\And {{F}_{2}}$ on any tangent of the ellipse.
Let us drawn a tangent at point P on the ellipse which we have shown in the below:

Now, we are going to find the foci ${{S}_{1}}\And {{S}_{2}}$ of the ellipse. We know that foci for the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to $\left( ae,0 \right)\And \left( -ae,0 \right)$. And here, “e” is the eccentricity of the ellipse: The formula for the eccentricity of the ellipse is as follows:
$\begin{aligned} & {{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}} \\\ & \Rightarrow {{e}^{2}}=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}} \\\ & \Rightarrow e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}} \\\ \end{aligned}$
Now, substituting the value of ${{a}^{2}}=5\And {{b}^{2}}=3$ in the above equation and we get,
$\begin{aligned} & e=\sqrt{\dfrac{5-3}{5}} \\\ & \Rightarrow e=\sqrt{\dfrac{2}{5}} \\\ \end{aligned}$
Substituting the above value of “e” in the foci coordinates we get,
$\begin{aligned} & {{S}_{1}}\left( \sqrt{5}\left( \sqrt{\dfrac{2}{5}} \right),0 \right)\And {{S}_{2}}\left( -\sqrt{5}\left( \sqrt{\dfrac{2}{5}} \right),0 \right) \\\ & \Rightarrow {{S}_{1}}\left( \sqrt{2},0 \right)\And {{S}_{2}}\left( -\sqrt{2},0 \right) \\\ \end{aligned}$
Now, plotting the above foci in the ellipse and then drawing perpendiculars from these foci to the tangent of the ellipse we get,

We know that equation of a tangent of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to:
$y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
Now, we can use the above equation of a tangent to find the tangent at point P for the given ellipse.
$y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
Substituting the value of ${{a}^{2}}=5\And {{b}^{2}}=3$ in the above equation and we get,
$y=mx\pm \sqrt{5{{m}^{2}}+3}$
Now, we are going to write the length of the perpendicular drawn from two foci to the tangent and we get,
To write the length of the perpendicular, first of all, we should write the above equation in the form of $ax+by+c=0$ so writing the above equation of tangent in this form we get,
$mx-y\pm \sqrt{5{{m}^{2}}+3}=0$
And we know that if we have to find the length of the perpendicular from the point say $\left( {{x}_{1}},{{y}_{1}} \right)$ on the line $ax+by+c=0$ it is calculated in the following way:
$\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Using the above formula we can find the length of ${{S}_{1}}{{F}_{1}}$ as follows:
$\begin{aligned} & \dfrac{\left| m\sqrt{2}+0\pm \sqrt{5{{m}^{2}}+3} \right|}{\sqrt{{{m}^{2}}+1}} \\\ & =\dfrac{\left( m\sqrt{2}\pm \sqrt{5{{m}^{2}}+3} \right)}{\sqrt{{{m}^{2}}+1}} \\\ \end{aligned}$
And length of ${{S}_{2}}{{F}_{2}}$ is also calculated in the above way and we get,

& \dfrac{\left| -m\sqrt{2}+0\pm \sqrt{5{{m}^{2}}+3} \right|}{\sqrt{{{m}^{2}}+1}} \\\ & =\dfrac{\left( -m\sqrt{2}\pm \sqrt{5{{m}^{2}}+3} \right)}{\sqrt{{{m}^{2}}+1}} \\\ \end{aligned}$$ Now, multiplying the above two lengths ${{S}_{1}}{{F}_{1}}\And {{S}_{2}}{{F}_{2}}$ we get, $=\dfrac{\left( m\sqrt{2}\pm \sqrt{5{{m}^{2}}+3} \right)}{\sqrt{{{m}^{2}}+1}}\dfrac{\left( -m\sqrt{2}\pm \sqrt{5{{m}^{2}}+3} \right)}{\sqrt{{{m}^{2}}+1}}$ The numerator in the above expression is the form of $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ so using this identity in the above we get, $\begin{aligned} & =\dfrac{\left( {{\left( \sqrt{5{{m}^{2}}+3} \right)}^{2}}-{{\left( m\sqrt{2} \right)}^{2}} \right)}{{{\left( \sqrt{{{m}^{2}}+1} \right)}^{2}}} \\\ & =\dfrac{5{{m}^{2}}+3-2{{m}^{2}}}{{{m}^{2}}+1} \\\ & =\dfrac{3{{m}^{2}}+3}{{{m}^{2}}+1} \\\ \end{aligned}$ Taking 3 as common in the numerator of the above expression we get, $\begin{aligned} & \dfrac{3\left( {{m}^{2}}+1 \right)}{{{m}^{2}}+1} \\\ & =3 \\\ \end{aligned}$ From the above, the result of $\left( {{S}_{1}}{{F}_{1}} \right)\left( {{S}_{2}}{{F}_{2}} \right)=3$. **So, the correct answer is “Option c”.** **Note:** There is a property which states that the product of the perpendicular drawn from the two foci on any tangent of the ellipse of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is equal to ${{b}^{2}}$ so in the above problem also, we are asked to find the product of the perpendicular drawn from the two foci on any tangent of the ellipse then its answer is equal to ${{b}^{2}}$ and the value of the ${{b}^{2}}$ in the given ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 3.