Question

The tangents drawn from a point P to the ellipse makes an angle ${{\theta }_{1}}\text{ and }{{\theta }_{2}}$with the major axis. Find the locus of P when:
${{\theta }_{1}}+{{\theta }_{2}}=2\alpha \left( \text{constant} \right)$

Answer 1

In the above question it is asked to find the locus of a point P which lies at the intersecting tangents of the ellipse. The tangents that touch the ellipse are the endpoints of the major axis. For this to work out we are first going to assume the x and y coordinate of P, then we will be using the tangent equation of ellipse and then substitute the assumed x and y coordinates to get a quadratic equation with use of quadratic equation we can simply use the above relation which is given as : ${{\theta }_{1}}+{{\theta }_{2}}=2\alpha \left( \text{constant} \right)$.

Complete step by step solution:
We know the ellipse equation which is stated as under:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
So, for first we are going to assume the x and y co-ordinates of P as h and k and we got:
$\Rightarrow p\left( x,y \right)\text{ }=\text{ }p\left( h,k \right)$
Now we also know the tangent equation for ellipse which is stated as below:
$y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
We know that this is the equation of tangent and point p lies on the tangent we can substitute the x and y co-ordinates i.e. h and k, in place of x and y in the equation and we get.

& \Rightarrow k=mh\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \\\ & \Rightarrow k-mh=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \\\ \end{aligned}$$ Now we will be squaring both the sides, from which we will get, $$\begin{aligned} & \Rightarrow {{\left( k-mh \right)}^{2}}={{\left( \pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \right)}^{2}} \\\ & \Rightarrow \left( {{k}^{2}}+{{m}^{2}}{{h}^{2}}-2mkh \right)={{a}^{2}}{{m}^{2}}+{{b}^{2}} \\\ & \Rightarrow \left( {{h}^{2}}-{{a}^{2}} \right){{m}^{2}}-2hkm+{{k}^{2}}-{{b}^{2}}=0......\left( 1 \right) \\\ \end{aligned}$$ Equation 1 became a quadratic equation and hence we can use the property of quadratic equation through which we will get: $$\begin{aligned} & {{m}_{1}}{{m}_{2}}=\dfrac{c}{a} \\\ & \Rightarrow {{m}_{1}}{{m}_{2}}=\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}}........\left( 2 \right) \\\ \end{aligned}$$ $$\begin{aligned} & {{m}_{1}}+{{m}_{2}}=-\dfrac{b}{a} \\\ & \Rightarrow {{m}_{1}}+{{m}_{2}}=\dfrac{2hk}{{{h}^{2}}-{{a}^{2}}}........\left( 3 \right) \\\ \end{aligned}$$ We also know that we can write slope as $$\tan \left( \theta \right)$$ so we can also say that: $${{m}_{1}}=\tan \left( {{\theta }_{1}} \right)$$ and $${{m}_{2}}=\tan \left( {{\theta }_{2}} \right)$$ Now from the relation that is mentioned in the question: $${{\theta }_{1}}+{{\theta }_{2}}=2\alpha \left( \text{constant} \right)$$ By taking tan on both sides we get: $$\begin{aligned} & \tan \left( {{\theta }_{1}}+{{\theta }_{2}} \right)=\tan \left( 2\alpha \right) \\\ & \Rightarrow \dfrac{\tan {{\theta }_{1}}+\tan {{\theta }_{2}}}{1-\tan {{\theta }_{1}}\tan {{\theta }_{2}}}=\tan \left( 2\alpha \right) \\\ & \Rightarrow \dfrac{{{m}_{1}}+{{m}_{2}}}{1-{{m}_{1}}{{m}_{2}}}=\tan \left( 2\alpha \right).....\left( 4 \right) \\\ \end{aligned}$$ Now from equation 2 and 3 we can substitute the respective values in equation 4 and we get $$\begin{aligned} & \Rightarrow \dfrac{\dfrac{2hk}{{{h}^{2}}-{{a}^{2}}}}{1-\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}}}=\tan \left( 2\alpha \right) \\\ & \Rightarrow \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}-{{k}^{2}}+{{b}^{2}}}=\tan \left( 2\alpha \right) \\\ \end{aligned}$$ Now we can write tan(2α) as $$\tan (2\alpha )=\dfrac{1}{\cot (2\alpha )}$$ And by substituting this in the above equation we will get: $$\begin{aligned} & \Rightarrow \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}-{{k}^{2}}+{{b}^{2}}}=\dfrac{1}{\cot (2\alpha )} \\\ & \Rightarrow 2hk\cot (2\alpha )={{h}^{2}}-{{a}^{2}}-{{k}^{2}}+{{b}^{2}} \\\ & \Rightarrow 2hk\cot (2\alpha )={{h}^{2}}-{{k}^{2}}+{{b}^{2}}-{{a}^{2}} \\\ \end{aligned}$$ Now we will substitute x and y with h and k respectively and we will get the locus of the point as: $$2xy\cot (2\alpha )={{x}^{2}}-{{y}^{2}}+{{b}^{2}}-{{a}^{2}}$$ So the locus of the point P when the relation $${{\theta }_{1}}+{{\theta }_{2}}=2\alpha \left( \text{constant} \right)$$ we get it as $$2xy\cot (2\alpha )={{x}^{2}}-{{y}^{2}}+{{b}^{2}}-{{a}^{2}}$$ **Note:** In the type of question that is mentioned above the main problem that arises is at the point on where to use the relation let us take example of this question, the relation that was stated was between angles, with those angles we can find the value of slope as we know that slope is always given in the tan type, so try to remember what does each trigonometric relation states and what does they signify.